NCERT Solutions For Class 11 Maths Chapter 4

NCERT Solutions Class 11 Maths Principle of Mathematical Induction

Ncert Solutions For Class 11 Maths Chapter 4 PDF Download

NCERT solutions for class 11 maths chapter 4 principles of mathematical induction is provide here so that students can refer to these solutions when they are facing difficulties while solving the questions. The principle of Mathematical Induction is one of the important and fundamentals for mathematical thinking is logical reasoning. In this topic, students will learn how to prove or verify the certain results that are expressed in terms of ‘n’ with the help of a specific technique, termed as the principle of mathematical induction. Usually, the Mathematical induction is used to prove the statements of the set of natural numbers. Altogether, there is two main Principle of Mathematical Induction, they are as follows:

According to the first principle of mathematical induction- the given statement is correct for the first natural number that is, for n=1, p (1) is true.

The second principle of mathematical induction is strong induction, which is more powerful than the first principle.

Mathematical induction is an important chapter for class 11 maths. Principles of mathematical induction hold higher weightage in any mathematical examination. To score good marks in class 11 maths examination one must solve the NCERT questions of each and every chapter. The questions provided in the NCERT books are prepared by subject experts so that students can practice the questions of different difficulty level.

NCERT Solutions, for class 11, chapter – 4 – Maths Principle of Mathematical Induction help students in preparing for their CBSE class 11 exams. Along with this, it also provides a basic foundation for their higher classes. These NCERT solutions are prepared by the team of our subject experts and are framed according to the NCERT and CBSE curriculum. For more detailed information, students can check with the below-given pdf files.

NCERT Solutions Class 11 Maths Chapter 4 Exercises

Exercise 4.1

Prove the following through principle of mathematical induction for all values of n, where n is a natural number.

1) \(1 + 3 + 3^{2} + …. + 3^{n – 1} = \frac{\left ( 3^{n} – 1 \right )}{2}\)

 

Sol:

The given statement is:

P(n) : \(1 + 3 + 3^{2} + …. + 3^{n – 1} = \frac{\left ( 3^{n} – 1 \right )}{2}\)

Now, for n = 1

P(1) = \(\frac{\left ( 3^{1} – 1 \right )}{2}\) = \(\frac{\left ( 3 – 1 \right )}{2}\)= \(\frac{2}{2}\)= 1

Thus, the P(n) is true for n=1

Let,\([ 2( k + 1) + 7 ]=[(2k + 7)+ 2]\)\([ 2( k + 1) + 7 ]=[(2k + 7)+ 2]\)\([ 2( k + 1) + 7 ]=[(2k + 7)+ 2]\)

P(k) be true, where k is a positive integer.

\(1 + 3 + 3^{2} + ….+ 3^{k – 1} = \frac{\left ( 3^{k} – 1 \right )}{2}\) . . . . . . . . . . (1)

Now, we will prove that P(k+1) is also true:

P(k + 1):

=\(\\1 + 3 + 3^{2} + ….+ 3^{k – 1} + 3^{\left (k + 1 \right ) – 1}\\\)

=\(\\\left (1 + 3 + 3^{2} + ….+ 3^{k – 1} \right )+ 3^{k}\\\)        [Using equation (1)]

=\(\\\frac{\left (3^{k} – 1 \right )}{2} + 3^{k}\\\)

=\(\\\frac{\left (3^{k} – 1 \right )+ 2.3^{k}}{2}\\\)

=\(\\\frac{(1 + 2)3^{k} – 1}{2}\\\)

=\(\\\frac{3.3^{k} – 1}{2}\\\)

=\(\\\frac{3^{k + 1} – 1}{2}\)

Thus, whenever P(k) proves to be true, P(k+1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

2: \(1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, n^{3}\) = \(\left ( \frac{n\;(n+1)}{2} \right )^{2}\)

 

Sol:

The given statement is:

P(n): \(1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, n^{3}\) = \(\left ( \frac{n\;(n+1)}{2} \right )^{2}\)

Now, for n = 1

P(1): \(1^{3} = 1 = \left ( \frac{1\left ( 1 + 1 \right )}{2} \right )^{2}\)= \(\left (\frac{1\times 2}{2} \right )^{2}\)= \(1^{2}\) = 1

Thus, the P(n) is true for n = 1

Let, P(k) be true, where ‘k’ is a positive integer.

\(1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, k^{3} = \left ( \frac{k(k+1)}{2} \right )^{2}\) . . . . . . . . . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true.

P(k + 1):

=\(\\1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, k^{3} + \left (k + 1 \right )^{3}\\\)

=\(\\\left (1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, k^{3} \right ) + \left (k + 1 \right )^{3}\\\)

=\(\\\left ( \frac{k\left ( k + 1 \right )}{2} \right )^{2} +\left ( k + 1 \right )^{3}\\\)    [From equation (1)]

=\(\\\frac{k^{2} \left ( k + 1 \right )^{2}}{4} + \left ( k + 1 \right )^{3}\\\)

=\(\\\frac{k^{2} \left ( k + 1 \right )^{2} + 4\left ( k + 1 \right )^{3}}{4}\\\)

=\(\\\frac{ \left ( k + 1 \right )^{2}\left \{ k^{2} + 4\left ( k + 1 \right ) \right \}}{4}\\\)

=\(\\\frac{ \left ( k + 1 \right )^{2}\left \{ k^{2} + 4k + 4 \right \}}{4}\\\)

=\(\\\frac{ \left ( k + 1 \right )^{2} \left ( k + 2 \right )^{2}}{4}\\\)

=\(\\\frac{ \left ( k + 1 \right )^{2} \left ( k + 1 + 1 \right )^{2}}{4}\\\)

=\(\\\left [\frac{ \left ( k + 1 \right ) \left ( k + 1 + 1 \right )}{2} \right ]^{2}\\\)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

3: \(1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + n } = \frac{2n}{n + 1}\)

 

Sol:

The given statement is:

P(n): \(1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + n } = \frac{2n}{n + 1}\)

Now, for n = 1

P(1): \(1 = \frac{2 \times 1}{1 + 1} = \frac{2}{2} = 1\)

Thus, the P(n) is true for n=1

Let, P(k) be true, where k is a positive integer.

\(1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + k } = \frac{2k}{k + 1}\) . . . . . . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true.

=\(\\1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + k } + \frac{1}{1 + 2 + 3 +… + k + \left ( k + 1 \right )}\\\)

=\(\\\left [1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + k } \right ]+ \frac{1}{1 + 2 + 3 +… + k + \left ( k + 1 \right )}\\\)

=\(\\\frac{2k}{k + 1}+ \frac{1}{1 + 2 + 3 +… + k + \left ( k + 1 \right )}\\\)      [From equation (1)]

= \(\\\frac{2k}{k + 1}+ \frac{1}{\left (\frac{\left ( k + 1 \right )\left ( k + 1 + 1 \right )}{2} \right )}\)            (\(\;1 + 2 + 3 + … + n = \frac{n\left ( n + 1 \right )}{2}\;\\\))

= \(\\\frac{2k}{k + 1}+ \frac{2}{{\left ( k + 1 \right )\left ( k + 2 \right )}}\\\)

= \(\\\frac{2}{\left (k + 1 \right )} \left ( k + \frac{1}{k + 2} \right )\\\)

= \(\\\frac{2}{\left (k + 1 \right )} \left ( \frac{k^{2} + 2k + 1 }{k + 2}\right )\\\)

= \(\\\frac{2}{\left (k + 1 \right )} \left ( \frac{\left (k + 1 \right ) ^{2}}{k + 2}\right )\\\)

= \(\\\frac{2\left ( k + 1 \right )}{\left ( k + 2 \right )}\\\)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

4: \(1.2.3 + 2.3.4 + … + n\left ( n + 1 \right )\left ( n + 2 \right ) = \frac{n \left ( n + 1 \right ) \left ( n + 2 \right )\left ( n + 3 \right )}{4}\)

 

Sol:

The given statement is:

P(n): \(1.2.3 + 2.3.4 + … + n\left ( n + 1 \right )\left ( n + 2 \right ) = \frac{n \left ( n + 1 \right ) \left ( n + 2 \right )\left ( n + 3 \right )}{4}\)

Now, for n = 1

P(1): \(1.2.3 = 6\)= \(\frac{1 \left ( 1 + 1 \right ) \left ( 1 + 2 \right )\left ( 1 + 3 \right )}{4}\)= \(\frac{1.2.3.4}{4}\)= 6

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

\(1.2.3 + 2.3.4 + … + k\left ( k + 1 \right )\left ( k + 2 \right ) = \frac{k \left ( k + 1 \right ) \left ( k + 2 \right )\left ( k + 3 \right )}{4}\)  . . . . . . . . . . (1)

Now, we will prove P(k + 1) is true.

=\(\\1.2.3 + 2.3.4 + … + k\left ( k + 1 \right )\left ( k + 2 \right ) + \left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3 \right )\\\)

=\([1.2.3 + 2.3.4 + … + k( k + 1)( k + 2 ) ] +n (k + 1)( k + 2 )( k + 3 )\\\)     [By using equation (1)]

=\(\\\frac{k \left ( k + 1 \right ) \left ( k + 2 \right )\left ( k + 3 \right )}{4} + \left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3\right )\\\)

Now, by using equation (1) :

=\(\\\left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3\right )\left ( \frac{k}{4} + 1\right )\\\)

=\(\\\frac{\left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3\right )\left ( k + 4 \right )}{4}\\\)

=\(\\\frac{\left ( k + 1 \right )\left ( k + 1 + 1 \right )\left ( k + 1 + 2\right )\left ( k + 1 + 3 \right )}{4}\\\)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

5: \(1.3 + 2. 3^{2}+ 3.3^{3} +…+ n. 3^{n} = \frac{(2n-1)3^{n+1}\, +\, 3}{4}\)

 

Sol:

The given statement is:

P(n): \(1.3 + 2. 3^{2}+ 3.3^{3} +…+ n. 3^{n} = \frac{(2n-1)3^{n+1}\, +\, 3}{4}\)

Now, for n = 1:

=\(\\\frac{\left ( 2.1 – 1 \right )3^{1+1} + 3}{4}\)= \(\frac{3^{2} + 3}{4}\)= \(\frac{12}{4}\)= 3

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

1.3 + 2. 3^{2}+ 3.3^{3} +…+ k. 3^{k} = \frac{(2k-1)3^{k+1}\, +\, 3}{4} . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=\(\\1.3 + 2.3^{2} + 3.3^{3} + … + k.3^{k} + \left ( k + 1 \right ).3^{k+1}\\\)

=\(\\\frac{\left ( 2k -1 \right )3^{k+1}+ 3}{4} + \left ( k + 1 \right )3^{k + 1}\)     [By using equation (1)]

=\(\\\frac{\left ( 2k -1 \right )3^{k+1}+ 3 + 4\left ( k + 1 \right )3^{k + 1}}{4}\\\)

= \(\\\frac{3^{k + 1}\left \{ 2k – 1 + 4\left ( k + 1 \right ) \right \} + 3}{4}\\\)

=\(\\\frac{3^{k + 1}\left \{ 6k + 3 \right \} + 3}{4}\\\)

=\(\\\frac{3^{k + 1}.3\left \{ 2k + 1 \right \} + 3}{4}\\\)

= \(\\\frac{3^{\left (k + 1 \right )+ 1}\left \{ 2k + 1 \right \} + 3}{4}\\\)

=\(\\\frac{\left \{ 2\left ( k + 1 \right )- 1 \right \}3^{\left ( k + 1 \right ) + 1} + 3}{4}\\\)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

6: \(1.2 + 2.3 + 3.4 + … + n.\left ( n + 1 \right ) = \left [ \frac{n\left ( n + 1 \right )\left ( n + 2 \right )}{3} \right ]\)

 

Sol:

The given statement is:

P(n): \(1.2 + 2.3 + 3.4 + … + n.\left ( n + 1 \right ) = \left [ \frac{n\left ( n + 1 \right )\left ( n + 2 \right )}{3} \right ]\)

Now, for n = 1:

=\(\frac{1\left ( 1 + 1 \right )\left ( 1 + 2 \right )}{3}\)= \(\frac{1.2.3}{3}\)= 2

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

\(1.2 + 2.3 + 3.4 + … + k.\left ( k + 1 \right ) = \left [ \frac{k\left ( k + 1 \right )\left ( k + 2 \right )}{3} \right ]\)  . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=\(\\\left [1.2 + 2.3 + 3.4 + … + k.\left ( k + 1 \right ) \right ] + \left ( k + 1 \right )\left ( k + 2 \right )\\\)

=\(\\\frac{k\left ( k + 1 \right )\left ( k + 2 \right )}{3} + \left ( k + 1 \right )\left ( k + 2 \right )\\\)

Now, by using equation (1):

=\(\\\left ( k + 1 \right )\left ( k + 2 \right )\left ( \frac{k}{3} + 1 \right )\\\)

=\(\\\frac{\left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3 \right )}{3}\\\)

=\(\\\frac{\left ( k + 1 \right )\left ( k + 1 + 1 \right )\left ( k + 1 + 2 \right )}{3}\\\)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

7: \(1.3 + 3.5 + 5.7 + … + \left ( 2n – 1 \right )\left ( 2n + 1 \right ) = \frac{n\left ( 4n^{2} + 6n – 1 \right )}{3}\)

 

Sol:

The given statement is:

\(1.3 + 3.5 + 5.7 + … + \left ( 2n – 1 \right )\left ( 2n + 1 \right ) = \frac{n\left ( 4n^{2} + 6n – 1 \right )}{3}\\\)

Now, for n = 1:

\(\\\frac{1\left ( 4.1^{2} + 6.1 – 1 \right )}{3}\)= \(\frac{4 + 6 – 1}{3}\)= \(\frac{9}{3}\) = 3

Thus, the P(n) is true for n = 1

Let, P(k) be true, where k is a positive integer.

\(1.3 + 3.5 + 5.7 + … + \left ( 2k – 1 \right )\left ( 2k + 1 \right ) = \frac{k\left ( 4k^{2} + 6k – 1 \right )}{3}\) . . . . . . . . .  (1)

Now we will prove P(k + 1) is also true:

=\(\\1.3 + 3.5 + 5.7 + … + \left ( 2k – 1 \right )\left ( 2k + 1 \right ) + \left \{ 2\left ( k + 1 \right ) – 1 \right \}\left \{ 2\left ( k + 1 \right ) + 1\right \}\\\)

=\(\\\frac{k\left ( 4k^{2} + 6k – 1\right )}{3} + \left ( 2k + 2 – 1 \right )\left ( 2k + 2 + 1 \right )\\\)

Now, by using equation (1):

=\(\\\frac{k\left ( 4k^{2} + 6k – 1\right )}{3} + \left ( 2k + 1 \right )\left ( 2k + 3 \right )\\\)

=\(\\\frac{k\left ( 4k^{2} + 6k – 1\right )}{3} + \left ( 4k^{2} + 8k + 3 \right )\\\)

=\(\\\frac{k\left ( 4k^{2} + 6k – 1\right )+ 3\left ( 4k^{2} + 8k + 3 \right )}{3}\\\)

=\(\\\frac{4k^{3} + 6k^{2} – k + 12k^{2} + 24k + 9}{3}\\\)

=\(\\\frac{4k^{3} + 18k^{2} + 23k + 9}{3}\\\)

=\(\\\frac{4k^{3} + 14k^{2} + 9k + 4k^{2} + 14k + 9}{3}\\\)

=\(\\\frac{k\left (4k^{2} + 14k + 9 \right )+ 1\left (4k^{2} + 14k + 9 \right )}{3}\\\)

=\(\\\frac{\left (k + 1 \right )\left (4k^{2} + 14k + 9 \right )}{3}\\\)

=\(\\\frac{\left (k + 1 \right )\left (4k^{2} + 8k + 4 + 6k + 6 – 1 \right )}{3}\\\)

=\(\\\frac{\left ( k + 1 \right )\left \{ 4\left ( k^{2} + 2k + 1\right ) + 6\left ( k + 1 \right ) – 1\right \}}{3}\\\)

=\(\\\frac{\left ( k + 1 \right )\left \{ 4\left ( k + 1\right )^{2} + 6\left ( k + 1 \right ) – 1\right \}}{3}\\\)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

8: \(1.2 + 2.2^{2} + 3.2^{2} + … + n.2^{n} = \left ( n – 1 \right )2^{n + 1} + 2\)

 

Sol:

The given statement is:

P(n): \(1.2 + 2.2^{2} + 3.2^{2} + … + n.2^{n} = \left ( n – 1 \right )2^{n + 1} + 2\)

Now, for n = 1:

=\(\left ( 1 – 1 \right )2^{1 + 1} + 2\) = 0 + 2= 2

Thus, the P(n) is true for n=1.

Let P(k) be true, where k is a positive integer:

\(\\1.2 + 2.2^{2} + 3.2^{2} + … + k.2^{k} = \left ( k – 1 \right )2^{k + 1} + 2\) . . . . . . . . (1)

Now we will prove P(k + 1) is also true:

=\([ 1.2 + 2.2^{2} + 3.2^{2} + … + k.2^{k} ] + ( k + 1 ). 2^{k + 1}\\\)

=\(\\\left ( k – 1 \right )2^{k + 1} + 2 + \left ( k + 1 \right ).2^{k + 1}\\\)

=\(\\2^{k + 1}\left \{ \left ( k – 1 \right ) + \left ( k + 1 \right ) \right \} + 2\\\)

=\(\\2^{k + 1}.2k + 2\\\)

=\(\\k.2^{\left ( k + 1 \right ) + 1} + 2\\\)

=\(\\\left \{ \left ( k + 1 \right ) – 1 \right \}2^{\left ( k + 1 \right ) + 1} + 2\\\)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

9: \(\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{n}} = 1 – \frac{1}{2^{n}}\)

 

Sol:

The given statement is:

\(\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{n}} = 1 – \frac{1}{2^{n}}\\\)

Now, for n = 1:

=\(1 – \frac{1}{2^{1}}\)= \(\frac{1}{2}\)

Thus, the P (n) is true for n = 1.

Let, P (k) be true, where k is a positive integer:

\(\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{k}} = 1 – \frac{1}{2^{k}}\) . . . . . . (1)

Now, we will prove P (k + 1) is also true:

=\(\\\left (\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{k}} \right ) + \frac{1}{2^{k + 1}}\\\)

=\(\\\left (1 – \frac{1}{2^{k}} \right ) + \frac{1}{2^{k + 1}}\\\)

Now, by using equation (1):

=\(\\1 – \frac{1}{2^{k}} + \frac{1}{2^{1}.2^{k}}\\\)

=\(\\1 – \frac{1}{2^{k}}\left (1 – \frac{1}{2} \right )\\\)

=\(\\1 – \frac{1}{2^{k}}\left ( \frac{1}{2} \right )\\\)

=\(\\1 – \frac{1}{2^{k + 1}}\\\)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

 10: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

\(\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}\)

 

Solution:

The given Statement is:
Q(n): \(\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}\)

Now, for n = 1

\(\\Q (1) = \frac{1}{2\cdot 5} = \frac{1}{(6(1)+4)} = \frac{1}{2\cdot 5}\\\)

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p):

= \(\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3p-1)(3p+2)} = \frac{p}{(6p+4)}\) . . . . . . . . . . . . . . (1)

Now, we have to  prove that Q(p+1) is also true.

Since, Q (p) is true, we have:

Q (p+1):

= \(\\\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3p-1)(3p+2)} + \frac{1}{[3(p + 1)-1][3(p + 1)+2]}\\\)

Now, by using equation (1):

=\(\\\frac{p}{(6p + 4)} + \frac{1}{[3p + 3 – 1]\;[3p + 3 + 2]}\\\)

=\(\\\frac{p}{(6p + 4)} + \frac{1}{(3p + 2)\;(3p + 3 + 2)}\\\)

=\(\\\frac{p}{2(3p + 2)} + \frac{1}{[3p + 3 – 1]\;[3p + 3 + 2]}\\\)

=\(\\\frac{p}{2(3p + 2)} + \frac{1}{[3p + 3 – 1]\;[3p + 3 + 2]}\\\)

=\(\\\frac{1}{2(3p + 2)}\left [ \frac{p}{2} +\frac{1}{(3p + 5)} \right ]\\\)

=\(\\\frac{1}{2(3p + 2)}\left [\frac{p(3p + 5) + 2}{(3p + 5)} \right ]\\\)

=\(\\\frac{1}{2(3p + 2)}\left [\frac{3p^{2} + 5p + 2}{(3p + 5)} \right ]\\\)

=\(\\\frac{1}{2(3p + 2)}\left [\frac{(3p + 2)(p + 1)}{(3p + 5)} \right ]\\\)

=\(\\\frac{p + 1}{6p + 10}\\\)

=\(\\\frac{p + 1}{6(p + 1) + 4}\\\)

Thus, whenever Q (p) proves to be true, Q (p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

 

 

11: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

\(\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot 4\cdot 5} + …….+ \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}\)

 

Solution:

The given statement is:

Q(n): \(\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot 4\cdot 5} + …….+ \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}\)

Now, for n = 1:

\(Q(1) = \frac{1}{1\cdot 2\cdot 3} = \frac{(1)((1)+3)}{4((1)+1)((1)+2)} = \frac{1}{1\cdot 2\cdot 3}\\\)

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number:

Q (p):

=\(\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot \cdot 4\cdot 5} + …….+ \frac{1}{p(p + 1)(p + 2)} = \frac{p(p + 3)}{4(p + 1)(p+2)}\). . . . . . . . . . . (1)

Now, we have to prove that Q (p+1) is also true.

Since, Q (p) is true, we have:

Q (p + 1):

=\(\\\frac{p(p + 3)}{4(p + 1)(p + 2)} + \frac{1}{(p + 1)(p + 2)(p + 3)}\\\)

Now, using equation (1):

=\(\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p + 3)}{4} + \frac{1}{(p + 3)} \right ]\\\)

=\(\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p + 3)^{2} + 4}{4(p + 3)} \right ]\\\)

=\(\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p^{2} + 6p + 9) + 4}{4(p + 3)} \right ]\\\)

=\(\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p^{2} + 6p + 9) + 4}{4(p + 3)} \right ]\\\)

=\(\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p^{3} + 6p^{2} + 9p + 4}{4(p + 3)} \right ]\\\)

=\(\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p^{2} + 2p + 1) + 4(p^{2} + 2p + 1)}{4(p + 3)} \right ]\\\)

=\(\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p + 1)^{2} + 4(p + 1)^{2}}{4(p + 3)} \right ]\\\)

=\(\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{(p + 1)^{2}(p + 4)}{4(p + 3)} \right ]\\\)

=\(\\\frac{(p + 1)[(p + 1) + 3]}{4[(p + 1) + 1][(p + 1)+ 2]}\\\)

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

 

 

12: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

\(a + ar + ar^{2} + ……. + ar^{n-1} = \frac{a(r^{n}-1)}{r-1}\)

 

Solution:

The given statement is:

Q(n): \(a + ar + ar^{2} + ……. + ar^{n-1} = \frac{a(r^{n}-1)}{r-1}\)

Now, for n = 1

\(Q(1) = a = \frac{a(r^{(1)}-1)}{r-1} = a\\\)

Thus, Q (1) proves to be true.

Let’s assume Q (p) is true, where p is a natural number.

Q (p)  = \(a + ar + ar^{2} + ……. + ar^{p-1} = \frac{a(r^{p}-1)}{r-1}\) . . . . . . . . (1)

Now, we have to prove that Q (p+1) is also true.

Since, Q (p) is true, we have:

Q(p+1) = \(a + ar + ar^{2} + ……. + ar^{p-1} + ar^{(p + 1)-1}\)

Now, using Equation (1):

=\(\\\frac{a(r^{p} – 1)}{r – 1} + ar^{p}\\\)

=\(\\\frac{a(r^{p} – 1) + ar^{p}(r – 1)}{r – 1}\\\)

=\(\\\frac{a(r^{p} – 1) + ar^{p}(r – 1) – ar^{p}}{r – 1}\\\)

=\(\\\frac{ar^{p} – a + ar^{p + 1} – ar^{p}}{r – 1}\\\)

=\(\\\frac{ar^{p + 1} – a}{r – 1}\\\)

=\(\\\frac{a(r^{p + 1} – 1)}{r – 1}\\\)

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

 

 

13: Provide a proof for the following with the help of mathematical induction principle for all values of n, where it is a natural number.

\(\left ( 1\, +\, \frac{3}{1} \right )\left ( 1\, +\, \frac{5}{4} \right )\left ( 1\, +\, \frac{7}{9} \right )……. \left ( 1\, +\, \frac{(2n+1)}{ n^{2}} \right ) = \left ( n+1 \right )^{2}\)

 

Solution:

The given statement is:

Q(n): \(\left ( 1\, +\, \frac{3}{1} \right )\left ( 1\, +\, \frac{5}{4} \right )\left ( 1\, +\, \frac{7}{9} \right )……. \left ( 1\, +\, \frac{(2n+1)}{ n^{2}} \right ) = \left ( n+1 \right )^{2}\)

Now, for n = 1:

\(Q(1) = \left ( 1\, +\, \frac{3}{1} \right ) = 4\\\)

\(\\\Rightarrow \left ( n+1 \right )^{2} = (1 + 1)^{2} = 4\\\)

Thus, Q (1) proves to be true.

Let’s assume Q (p) is true, where p is a natural number.

Q (p):

=\(\left ( 1\, +\, \frac{3}{1} \right )\left ( 1\, +\, \frac{5}{4} \right )\left ( 1\, +\, \frac{7}{9} \right )……. \left ( 1\, +\, \frac{(2p+1)}{ p^{2}} \right ) = \left ( p+1 \right )^{2}\) . . . . . . . . . . . (1)

Now we have to prove that Q(p+1) is also true.

Since Q (p) is true, we have:

Q(p+1):

=\(\left ( 1\, +\, \frac{3}{1} \right )\left ( 1\, +\, \frac{5}{4} \right )\left ( 1\, +\, \frac{7}{9} \right )……. \left ( 1\, +\, \frac{(2p+1)}{ p^{2}} \right )\left ( 1\, +\, \frac{(2(p+1) + 1}{ (p + 1)^{2}} \right )\\\)

Now, using equation (1):

=\(\\(p + 1)^{2}\left ( 1\, +\, \frac{(2(p+1) + 1}{ (p + 1)^{2}} \right )\\\)

=\(\\(p + 1)^{2}\left ( \frac{(p+1)^{2} + 2(p+1) + 1}{ (p + 1)^{2}} \right )\\\)

\(\\\Rightarrow\) \((p+1)^{2} + 2(p+1) + 1= \left [ (p + 1) + 1 \right ]^{2}\\\)
Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

 

 

14: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

\(\left ( 1\, +\, \frac{1}{1} \right )\left ( 1\, +\, \frac{1}{2} \right )\left ( 1\, +\, \frac{1}{3} \right )…….\left ( 1\, +\, \frac{1}{n} \right ) = (n+1)\)

 

Solution:

The given statement is:

Q(n): \(\left ( 1\, +\, \frac{1}{1} \right )\left ( 1\, +\, \frac{1}{2} \right )\left ( 1\, +\, \frac{1}{3} \right )…….\left ( 1\, +\, \frac{1}{n} \right ) = (n+1)\)

Now, for n = 1:

\(Q(1) = \left ( 1\, +\, \frac{1}{1} \right ) = 2 \\ \Rightarrow (n+1) = 1 + 1 = 2\\\)

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p):

=\(\left ( 1\, +\, \frac{1}{1} \right )\left ( 1\, +\, \frac{1}{2} \right )\left ( 1\, +\, \frac{1}{3} \right )…….\left ( 1\, +\, \frac{1}{p} \right ) = (p+1)\). . . . .(1)

Now, we have to prove that Q (p+1) is also true.

Since, Q (p) is true, we have:

Q (p+1):

=\(\\\left ( 1\, +\, \frac{1}{1} \right )\left ( 1\, +\, \frac{1}{2} \right )\left ( 1\, +\, \frac{1}{3} \right )…….\left ( 1\, +\, \frac{1}{p} \right )\left ( 1\, +\, \frac{1}{p + 1} \right )\\\)

Now, using Equation (1):

=\(\\(p + 1)\left ( 1\, +\, \frac{1}{p + 1} \right )\)\(= (p + 1)\left (\frac{(p + 1) + 1}{p + 1} \right )\\\)

= (p + 1) + 1

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

 

 

15: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

\(1^{2} + 3^{2} + 5^{2} +. …..+ (2n-1)^{2} = \frac{n(2n-1)(2n+1)}{3}\)

 

Solution:

The given statement is:

Q(n): \(1^{2} + 3^{2} + 5^{2} +. …..+ (2n-1)^{2} = \frac{n(2n-1)(2n+1)}{3}\)

Now, for n = 1:

\(Q(1) = 1^{2} = 1\;and\; \frac{(1)(2(1)-1)(2(1)+1)}{3} = \frac{3}{3} = 1\\\)

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q (p):

=\(1^{2} + 3^{2} + 5^{2} +. …..+ (2p-1)^{2} = \frac{p(2p-1)(2p+1)}{3}\) . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q(p) is true, we have:

Q (p+1):

=\(\\1^{2} + 2^{2} + 3^{2} +……+ (2k – 1)^{2}] + (2\;(p + 1) – 1)^{2}\\\)

=\(\\\frac{p(2p – 1)(2p + 1)}{3} + (2p + 2 -1)^{2}\\\)

=\(\\\frac{p(2p – 1)(2p + 1)}{3} + (2p + 1)^{2}\\\)

=\(\\\frac{p(2p – 1)(2p + 1) + 3(2p + 1)^{2}}{3}\\\)

=\(\\\frac{(2p + 1)(p(2p – 1) + 3(2p + 1))}{3}\\\)

=\(\\\frac{(2p + 1)(2p^{2} – p + 6p + 3)}{3}\\\)

=\(\\\frac{(2p + 1)(2p^{2} + 5p + 3)}{3}\\\)

=\(\\\frac{(2p + 1)(2p^{2} + 2p + 3p + 3)}{3}\\\)

=\(\\\frac{(2p + 1)(2p(p + 1) + 3(p + 1))}{3}\\\)

=\(\\\frac{(2p + 1)(p + 1)(2p + 3)}{3}\\\)

=\(\\\frac{(p + 1)[2(p + 1) – 1][2(p + 1) + 1]}{3}\\\)

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

 

 

16: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

\(\frac{1}{1\cdot 4} + \frac{1}{4\cdot 7} + \frac{1}{7\cdot 10} + …… + \frac{1}{(3n-2)(3n+1} = \frac{n}{3n+1}\)

 

Solution:

The given statement is:

Q(n): \(\frac{1}{1\cdot 4} + \frac{1}{4\cdot 7} + \frac{1}{7\cdot 10} + …… + \frac{1}{(3n-2)(3n+1} = \frac{n}{3n+1}\)

Now, for n = 1:

\(Q(1) = \frac{1}{1.4} = \frac{1}{(3(1)-2)(3(1)+1)} = \frac{1}{1.4}\\\)

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p):

=\(\frac{1}{1\cdot 4} + \frac{1}{4\cdot 7} + \frac{1}{7\cdot 10} + …… + \frac{1}{(3p – 2)(3p + 1)} = \frac{p}{3p + 1}\) . . . . . . . . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q (p) is true, we have:

Q (p+1):

=\(\frac{1}{1\cdot 4} + \frac{1}{4\cdot 7} + \frac{1}{7\cdot 10} +…+ \frac{1}{(3p – 2)(3p + 1)} + \frac{1}{[3(p + 1) – 2][3(p + 1)+ 1]}\)

Now, using Equation (1):

=\(\\\frac{1}{(3p + 1)} + \frac{1}{(3p + 1)(3p + 4)]}\\\)

=\(\\\frac{1}{(3p + 1)}\left [ p + \frac{1}{(3p + 4)} \right ]\\\)

=\(\\\frac{1}{(3p + 1)}\left [ \frac{p(3p + 4) + 1}{(3p + 4)} \right ]\\\)

=\(\\\frac{1}{(3p + 1)}\left [ \frac{3p^{2} + 4p + 1}{(3p + 4)} \right ]\\\)

=\(\\\frac{1}{(3p + 1)}\left [ \frac{3p^{2} + 3p + p + 1}{(3p + 4)} \right ]\\\)

=\(\\\frac{1}{(3p + 1)}\left [ \frac{(3p + 1)(p + 1)}{(3p + 4)} \right ]\\\)

=\(\\\frac{(p + 1)}{3(p + 1) + 1}\\\)

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

 

 

17: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

\(\frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \frac{1}{7\cdot 9} + ……. + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}\)

 

Solution:

The given statement is:

Q(n): \(\frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \frac{1}{7\cdot 9} + ……. + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}\)

Now, for n = 1:

\(Q(1) = \frac{1}{3\cdot 5} = \frac{1}{3(2(1)+3)} = \frac{1}{3\cdot 5}\\\)

Thus, Q(1) proves to be true:

Let us assume that Q(p) is true, where p is a natural number.

Therefore, Q(p):

=\(\frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \frac{1}{7\cdot 9} + ……. + \frac{1}{(2p + 1)(2p + 3)} = \frac{p}{3(2p + 3)}\) . . . . . . . (1)

Now, we have to prove that Q (p+1) is also true.

Since, Q(p) is true, we have:

Q(p+1):

=\(\\\frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \frac{1}{7\cdot 9} + ……. + \frac{1}{(2p + 1)(2p + 3)} + \frac{1}{[2(p + 1) + 1][2(p + 1) + 3]}\\\)

Now, Using Equation (1):

=\(\\\frac{p}{3(2p + 3)} + \frac{1}{(2p + 3)(2p + 5)}\\\)

=\(\\\frac{1}{3(2p + 3)}\left [ \frac{p}{3} + \frac{1}{(2p + 5)} \right ]\\\)

=\(\\\frac{1}{3(2p + 3)}\left [\frac{p(2p + 5) + 3}{(2p + 5)} \right ]\\\)

=\(\\\frac{1}{3(2p + 3)}\left [\frac{2p^{2} + 5p + 3}{(2p + 5)} \right ]\\\)

=\(\\\frac{1}{3(2p + 3)}\left [\frac{2p^{2} + 2p + 3p + 3}{(2p + 5)} \right ]\\\)

=\(\\\frac{1}{3(2p + 3)}\left [\frac{2p(p + 1) + 3(p + 1)}{(2p + 5)} \right ]\\\)

=\(\\\frac{1}{3(2p + 3)}\left [\frac{(2p + 3)(p + 1)}{(2p + 5)} \right ]\\\)

=\(\\\frac{p + 1}{3[2(p + 1) + 3]}\\\)

Thus, whenever Q(p) proves to be true, Q(p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q(n) is true for all the natural numbers.

 

 

18: Provide a proof for the following through principle of mathematical induction for all values of n, where n is a natural number.

\(1 + 2 + 3 + …… + n < \frac{1}{8}(2n+1)^{2}\)

Solution:

The given statement is:

Q(n): \(1 + 2 + 3 + …… + n < \frac{1}{8}(2n+1)^{2}\)

Now, for n = 1:

\(Q(1) = 1 = \frac{1}{8}(2(1)+1)^{2} = \frac{9}{8} \\ \Rightarrow 1 < \frac{9}{8}\\\)

Thus, Q(1) proves to be true.

Let’s assume Q(p) is true, where p is a natural number.

Q(p) = \(1 + 2 + 3 + …… + p < \frac{1}{8}(2p+1)^{2}\) . . . . . . (1)

Now, we have to prove that Q(p+1) is also true.

Since, Q (p) is true, we have:

Now, Using Equation (1):

=\((1 + 2 + 3 + …… + p) + (p + 1) < \frac{1}{8}(2p+1)^{2} + (p + 1)\\\)

<\(\\\frac{1}{8}[(2p + 1)^{2} + 8(p + 1)]\\\)

<\(\\\frac{1}{8}[4p^{2} + 4p + 1 + 8p + 8]\\\)

<\(\\\frac{1}{8}[4p^{2} + 12p + 9]\\\)

<\(\\\frac{1}{8}[(2p +3)^{2}]\\\)

<\(\\\frac{1}{8}[(2(p + 1) + 1)^{2}]\\\)

Thus, \(\\(1 + 2 + 3 + …… + p) + (p + 1) < \frac{1}{8}(2p+1)^{2} + (p + 1)\)

Thus, whenever Q (p) proves to be true, Q (p+1) subsequently proves to be true.

Therefore, with the help of induction principle it can be proved that the statement Q (n) is true for all the natural numbers.

 

 

19: n (n + 1)(n + 5) is a multiple of 3.

 

Sol:

The given statement is:

P(n): n (n + 1) (n + 5) is a multiple of 3

Now, for = 1:

= 1(1 + 1)(1 + 5) = 12

Thus, the P(n) is true for n = 1.

Let, P(k) be true, where k is a positive integer.

k(k + 1)(k + 5) is a multiple of 3

Therefore, k(k + 1)(k + 5) = 3m, where m \(\in\) N . . . . . . . . (1) 

Now, we will prove P(k + 1) is also true.

=\(\\\left ( k + 1 \right )\left \{ \left ( k + 1 \right ) + 1 \right \}\left \{ \left ( k + 1 \right ) + 5\right \}\)

= \(\\\left ( k + 1 \right )\left ( k + 2 \right )\left \{ \left ( k + 5 \right ) + 1\right \}\)

= \(\\\left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 5 \right ) + \left ( k + 1 \right )\left ( k + 2 \right )\)

= \(\\\left \{ k \left ( k + 1 \right )\left ( k + 5 \right ) + 2\left ( k + 1 \right )\left ( k + 5 \right ) \right \} + (k + 1)(k + 2)\)

= \(\\3m + (k + 1)\left \{ 2(k + 5) + (k + 2) \right \}\)

= \(\\3m + (k + 1)\left \{ 2k + 10 + k + 2\right \}\)

= \(\\3m + (k + 1)\left \{ 3k + 12 \right \}\)

= \(\\3m + 3(k + 1)(k + 4)\)

= \(\\3\left \{ m + (k + 1)(k + 4) \right \}\)

= \(\\3 \times q\), where q = \(\left \{ m + (k + 1)(k + 4) \right \}\) \(\in\) N.

Therefore, \(\left ( k + 1 \right )\left \{ \left ( k + 1 \right ) + 1 \right \}\left \{ \left ( k + 1 \right ) + 5\right \}\) is a multiple of 3.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

20: \(10^{2n – 1} + 1\) is divisible by 11.

 

Sol:

The given statement is:

\(10^{2n – 1} + 1\) is divisible by 11

Now, for n = 1

=\(10^{2.1 – 1} + 1\) = 11

Thus, the P(n) is true for n = 1.

Then, P(k) is also true, where k is a positive integer.

\(10^{2k – 1} + 1\) is divisible by 11.

\(10^{2k – 1} + 1\) = 11m, where m \(\in\) N  . . . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=\(\\10^{2\left ( k + 1 \right )- 1} + 1\)

=\(\\10^{2k + 2 – 1} + 1\)

=\(\\10^{2k + 1} + 1\)

=\(\\10^{2}\left ( 10^{2k – 1} + 1 – 1 \right ) + 1\)

=\(\\10^{2}\left ( 10^{2k – 1} + 1 \right ) – 10^{2} + 1\)

=\(\\10^{2}. 11m – 100 + 1\)          [using equation (1)]

=\(\\100 \times 11m – 99\)

=\(\\11\left ( 100m – 9 \right )\)

= 11 r, where r = \(\left (100m – 9 \right )\) is some natural number.

Therefore,  \(10^{2\left ( k + 1 \right )- 1} + 1\) is divisible by 11.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

21. \(x^{2n} – y^{2n}\) is divisible by x + y.

 

Sol:

The given statement is:

P(n): \(x^{2n} – y^{2n}\) is divisible by x + y.

Now, for n = 1

= \(x^{2 \times 1} – y^{2 \times 1}\)= \(x^{2} – y^{2}\)= (x + y) (x – y)

Therefore, it is divisible by (x + y).

Thus, the P(n) is true for n=1.

Let, P(k) is also true, where k is a positive integer.

\(x^{2k} – y^{2k}\) is divisible by (x + y).

Let \(x^{2k} – y^{2k}\) = m (x + y), where m \(\in\) N . . . . (1)

Now, we will prove P(k + 1) is also true:

=\(\\x^{2\left ( k + 1 \right ) } – y^{2\left ( k + 1 \right )}\)

=\(\\x^{2k}.x^{2} – y^{2k}.y^{2}\)

=\(\\x^{2}\left ( x^{2k} – y^{2k} + y^{2k}\right ) – y^{2k}. y^{2}\)

=\(\\x^{2}\left \{ m \left ( x + y \right ) + y^{2k}\right \} – y^{2k}.y^{2}\)             [using equation (1)]

=\(\\m \left ( x + y \right )x^{2} + y^{2k}.x^{2} – y^{2k}.y^{2}\)

=\(\\m \left ( x + y \right )x^{2} + y^{2k}\left (x^{2} – y^{2} \right )\)

=\(\\m \left ( x + y \right )x^{2} + y^{2k}\left (x + y \right )\left ( x – y \right )\)

=\(\\\left ( x + y \right )\left \{ mx^{2} + y^{2k}\left ( x – y \right ) \right \}\), which is the factor of (x + y).

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

 

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

22: \(3^{2n + 2} – 8n – 9\) is divisible by 8.

 

Sol:

The given statement is:

P(n): \(3^{2n + 2} – 8n – 9\) is divisible by 8.

Now, for n = 1:

=\(3^{2.1 + 2} – 8.1 – 9\)= \(3^{4} – 17\)= 64

Therefore, it is divisible by 8.

Thus, the P(n) is true for n = 1.

Let, P(k) be true, where k is a positive integer.

\(3^{2k + 2} – 8k – 9\) is divisible by 8.

\(3^{2k + 2} – 8k – 9 = 8m\), where m \(\in\) N. . . . . . (1)

Now, we will prove P(k + 1) is also true:

=\(\\3^{2\left (k + 1 \right ) + 2} – 8\left ( k + 1 \right ) – 9\)

=\(\\3^{2k + 2}.3^{2} – 8k – 8 – 9\)

=\(\\3^{2} \left ( 3^{2k + 2} – 8k – 9 + 8k + 9 \right ) – 8k – 17\)

=\(\\3^{2} \left ( 3^{2k + 2} – 8k – 9 \right ) + 3^{2}\left ( 8k + 9 \right ) – 8k – 17\)

=\(\\9.8m + 9\left ( 8k + 9 \right ) – 8k – 17\)

=\(\\9.8m + 72k + 81 – 8k – 17\)

=\(\\9.8m + 64k + 64\)

=\(\\8\left (9m + 8k + 8 \right )\)

= 8r, where r = \(\left (9m + 8k + 8 \right )\) is a natural number.

Therefore, \(3^{2\left (k + 1 \right ) + 2} – 8\left ( k + 1 \right ) – 9\) is divisible by 8.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

23: \(41^{n} – 14^{n}\) is a multiple of 27.

 

Sol:

The given statement is:

P(n): \(41^{n} – 14^{n}\) is a multiple of 27

Now, for n = 1:

=\(41^{1} – 14^{1}\)= 27, which is the multiple of 27

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

\(41^{k} – 14^{k}\) is a multiple of 27.

\(41^{k} – 14^{k}\) = 27m , where m \(\in\) N . . . . . . . . (1)

Now, we will prove P(k + 1) also is true:

=\(\\41^{k + 1} – 14^{k + 1}\)

=\(\\41^{k}.41 – 14^{k}.14\)

=\(\\41 \left (41^{k} – 14^{k} + 14^{k} \right ) – 14^{k}.14\)

=\(\\41 \left (41^{k} – 14^{k} \right ) + 41.14^{k} – 14^{k}.14\)

=\(\\41.27m + 14^{k} \left ( 41 – 14 \right )\)

=\(\\41.27m + 27.14^{k}\)

=\(\\27 \left (41m + 14^{k} \right )\)

=\(\\27 \times r\), where r = \(\left (41m + 14^{k} \right )\) is a natural number

Therefore, \(41^{k + 1} – 14^{k + 1}\) is a multiple of 27.

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

 

 

24: \(\left ( 2n + 7 \right ) < \left ( n + 3 \right )^{2}\)

 

Sol:

The given statement is:

P(n): \(\left ( 2n + 7 \right ) < \left ( n + 3 \right )^{2}\)

Now, for n = 1:

=\(\left ( 2.1 + 7 \right ) < \left ( 1 + 3 \right )^{2}\)= \(\left ( 9 \right ) < \left ( 4 \right )^{2}\)= \(9 < 16\)

Thus, the P(n) is true for n=1.

Let, P(k) be true, where k is a positive integer.

\(\left (2k + 7 \right ) < \left ( k + 3 \right )^{2}\) . . . . . . . . (1)

Now, we will prove P(k + 1) is also true:

=\([ 2( k + 1) + 7 ]=[(2k + 7)+ 2]\\\)

=\(\\\left [ 2 \left ( k + 1 \right ) + 7 \right ]\) = \(\left (2k + 7 \right ) + 2\) < \(\left ( k + 3 \right )^{2} + 2\\\)     [using equation (1)]

=\(\\2\left ( k + 1 \right ) + 7 < k^{2} + 6k + 9 + 2\\\)

=\(\\2\left ( k + 1 \right ) + 7 < k^{2} + 6k + 11\\\)

Now,\(\\k^{2} + 6k + 11 < k^{2} + 8k + 16\\\)

\(\\2\left ( k + 1 \right ) + 7 < \left ( k + 4 \right )^{4}\\\)

Thus, whenever P(k) proves to be true, P(k + 1) subsequently proves to be true.

Therefore, with the help of mathematical induction principle it can be proved that the statement P(n) is true for all the natural numbers.

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