Chapter 4: Principle of Mathematical Induction

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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