# NCERT Solutions For Class 11 Maths Chapter 4

## NCERT Solutions Class 11 Maths Principle of Mathematical Induction

Mathematical induction is an important chapter for class 11 maths. Principles of mathematical induction holds 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 book are prepared by subject matter experts so that students can practice questions of different difficulty level.

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.

### 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+32+.+3n1=(3n1)2$1 + 3 + 3^{2} + …. + 3^{n – 1} = \frac{\left ( 3^{n} – 1 \right )}{2}$

Sol:

The given statement is:

P(n) : 1+3+32+.+3n1=(3n1)2$1 + 3 + 3^{2} + …. + 3^{n – 1} = \frac{\left ( 3^{n} – 1 \right )}{2}$

Now, for n = 1

P(1) = (311)2$\frac{\left ( 3^{1} – 1 \right )}{2}$ = (31)2$\frac{\left ( 3 – 1 \right )}{2}$= 22$\frac{2}{2}$= 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]$[ 2( k + 1) + 7 ]=[(2k + 7)+ 2]$[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+32+.+3k1=(3k1)2$1 + 3 + 3^{2} + ….+ 3^{k – 1} = \frac{\left ( 3^{k} – 1 \right )}{2}$ . . . . . . . . . . (1)

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

P(k + 1):

=1+3+32+.+3k1+3(k+1)1$\\1 + 3 + 3^{2} + ….+ 3^{k – 1} + 3^{\left (k + 1 \right ) – 1}\\$

=(1+3+32+.+3k1)+3k$\\\left (1 + 3 + 3^{2} + ….+ 3^{k – 1} \right )+ 3^{k}\\$        [Using equation (1)]

=(3k1)2+3k$\\\frac{\left (3^{k} – 1 \right )}{2} + 3^{k}\\$

=(3k1)+2.3k2$\\\frac{\left (3^{k} – 1 \right )+ 2.3^{k}}{2}\\$

=(1+2)3k12$\\\frac{(1 + 2)3^{k} – 1}{2}\\$

=3.3k12$\\\frac{3.3^{k} – 1}{2}\\$

=3k+112$\\\frac{3^{k + 1} – 1}{2}$

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: 13+23+33++n3$1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, n^{3}$ = (n(n+1)2)2$\left ( \frac{n\;(n+1)}{2} \right )^{2}$

Sol:

The given statement is:

P(n): 13+23+33++n3$1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, n^{3}$ = (n(n+1)2)2$\left ( \frac{n\;(n+1)}{2} \right )^{2}$

Now, for n = 1

P(1): 13=1=(1(1+1)2)2$1^{3} = 1 = \left ( \frac{1\left ( 1 + 1 \right )}{2} \right )^{2}$= (1×22)2$\left (\frac{1\times 2}{2} \right )^{2}$= 12$1^{2}$ = 1

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

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

13+23+33++k3=(k(k+1)2)2$1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, k^{3} = \left ( \frac{k(k+1)}{2} \right )^{2}$ . . . . . . . . . . . . . . . . . (1)

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

P(k + 1):

=13+23+33++k3+(k+1)3$\\1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, k^{3} + \left (k + 1 \right )^{3}\\$

=(13+23+33++k3)+(k+1)3$\\\left (1^{3}\, +\, 2^{3}\, +\, 3^{3}\, +\, ……\, +\, k^{3} \right ) + \left (k + 1 \right )^{3}\\$

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

=k2(k+1)24+(k+1)3$\\\frac{k^{2} \left ( k + 1 \right )^{2}}{4} + \left ( k + 1 \right )^{3}\\$

=k2(k+1)2+4(k+1)34$\\\frac{k^{2} \left ( k + 1 \right )^{2} + 4\left ( k + 1 \right )^{3}}{4}\\$

=(k+1)2{k2+4(k+1)}4$\\\frac{ \left ( k + 1 \right )^{2}\left \{ k^{2} + 4\left ( k + 1 \right ) \right \}}{4}\\$

=(k+1)2{k2+4k+4}4$\\\frac{ \left ( k + 1 \right )^{2}\left \{ k^{2} + 4k + 4 \right \}}{4}\\$

=(k+1)2(k+2)24$\\\frac{ \left ( k + 1 \right )^{2} \left ( k + 2 \right )^{2}}{4}\\$

=(k+1)2(k+1+1)24$\\\frac{ \left ( k + 1 \right )^{2} \left ( k + 1 + 1 \right )^{2}}{4}\\$

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

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+11+2+11+2+3+.+11+2+3++n=2nn+1$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + n } = \frac{2n}{n + 1}$

Sol:

The given statement is:

P(n): 1+11+2+11+2+3+.+11+2+3++n=2nn+1$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + n } = \frac{2n}{n + 1}$

Now, for n = 1

P(1): 1=2×11+1=22=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+11+2+11+2+3+.+11+2+3++k=2kk+1$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + k } = \frac{2k}{k + 1}$ . . . . . . . . . . . . . . (1)

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

=1+11+2+11+2+3+.+11+2+3++k+11+2+3++k+(k+1)$\\1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + k } + \frac{1}{1 + 2 + 3 +… + k + \left ( k + 1 \right )}\\$

=[1+11+2+11+2+3+.+11+2+3++k]+11+2+3++k+(k+1)$\\\left [1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + …. + \frac{1}{1 + 2 + 3 +… + k } \right ]+ \frac{1}{1 + 2 + 3 +… + k + \left ( k + 1 \right )}\\$

=2kk+1+11+2+3++k+(k+1)$\\\frac{2k}{k + 1}+ \frac{1}{1 + 2 + 3 +… + k + \left ( k + 1 \right )}\\$      [From equation (1)]

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

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

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

= 2(k+1)(k2+2k+1k+2)$\\\frac{2}{\left (k + 1 \right )} \left ( \frac{k^{2} + 2k + 1 }{k + 2}\right )\\$

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

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

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

Now, for n = 1

P(1): 1.2.3=6$1.2.3 = 6$= 1(1+1)(1+2)(1+3)4$\frac{1 \left ( 1 + 1 \right ) \left ( 1 + 2 \right )\left ( 1 + 3 \right )}{4}$= 1.2.3.44$\frac{1.2.3.4}{4}$= 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(k+1)(k+2)=k(k+1)(k+2)(k+3)4$1.2.3 + 2.3.4 + … + k\left ( k + 1 \right )\left ( k + 2 \right ) = \frac{k \left ( k + 1 \right ) \left ( k + 2 \right )\left ( k + 3 \right )}{4}$  . . . . . . . . . . (1)

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

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

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

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

Now, by using equation (1) :

=(k+1)(k+2)(k+3)(k4+1)$\\\left ( k + 1 \right )\left ( k + 2 \right )\left ( k + 3\right )\left ( \frac{k}{4} + 1\right )\\$

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

=(k+1)(k+1+1)(k+1+2)(k+1+3)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+2.32+3.33++n.3n=(2n1)3n+1+34$1.3 + 2. 3^{2}+ 3.3^{3} +…+ n. 3^{n} = \frac{(2n-1)3^{n+1}\, +\, 3}{4}$

Sol:

The given statement is:

P(n): 1.3+2.32+3.33++n.3n=(2n1)3n+1+34$1.3 + 2. 3^{2}+ 3.3^{3} +…+ n. 3^{n} = \frac{(2n-1)3^{n+1}\, +\, 3}{4}$

Now, for n = 1:

=(2.11)31+1+34$\\\frac{\left ( 2.1 – 1 \right )3^{1+1} + 3}{4}$= 32+34$\frac{3^{2} + 3}{4}$= 124$\frac{12}{4}$= 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+2.32+3.33++k.3k+(k+1).3k+1$\\1.3 + 2.3^{2} + 3.3^{3} + … + k.3^{k} + \left ( k + 1 \right ).3^{k+1}\\$

=(2k1)3k+1+34+(k+1)3k+1$\\\frac{\left ( 2k -1 \right )3^{k+1}+ 3}{4} + \left ( k + 1 \right )3^{k + 1}$     [By using equation (1)]

=(2k1)3k+1+3+4(k+1)3k+14$\\\frac{\left ( 2k -1 \right )3^{k+1}+ 3 + 4\left ( k + 1 \right )3^{k + 1}}{4}\\$

= 3k+1{2k1+4(k+1)}+34$\\\frac{3^{k + 1}\left \{ 2k – 1 + 4\left ( k + 1 \right ) \right \} + 3}{4}\\$

=3k+1{6k+3}+34$\\\frac{3^{k + 1}\left \{ 6k + 3 \right \} + 3}{4}\\$

=3k+1.3{2k+1}+34$\\\frac{3^{k + 1}.3\left \{ 2k + 1 \right \} + 3}{4}\\$

= 3(k+1)+1{2k+1}+34$\\\frac{3^{\left (k + 1 \right )+ 1}\left \{ 2k + 1 \right \} + 3}{4}\\$

={2(k+1)1}3(k+1)+1+34$\\\frac{\left \{ 2\left ( k + 1 \right )- 1 \right \}3^{\left ( k + 1 \right ) + 1} + 3}{4}\\$

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

Sol:

The given statement is:

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

Now, for n = 1:

=1(1+1)(1+2)3$\frac{1\left ( 1 + 1 \right )\left ( 1 + 2 \right )}{3}$= 1.2.33$\frac{1.2.3}{3}$= 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.(k+1)=[k(k+1)(k+2)3]$1.2 + 2.3 + 3.4 + … + k.\left ( k + 1 \right ) = \left [ \frac{k\left ( k + 1 \right )\left ( k + 2 \right )}{3} \right ]$  . . . . . . . . . (1)

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

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

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

Now, by using equation (1):

=(k+1)(k+2)(k3+1)$\\\left ( k + 1 \right )\left ( k + 2 \right )\left ( \frac{k}{3} + 1 \right )\\$

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

=(k+1)(k+1+1)(k+1+2)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++(2n1)(2n+1)=n(4n2+6n1)3$1.3 + 3.5 + 5.7 + … + \left ( 2n – 1 \right )\left ( 2n + 1 \right ) = \frac{n\left ( 4n^{2} + 6n – 1 \right )}{3}$

Sol:

The given statement is:

1.3+3.5+5.7++(2n1)(2n+1)=n(4n2+6n1)3$1.3 + 3.5 + 5.7 + … + \left ( 2n – 1 \right )\left ( 2n + 1 \right ) = \frac{n\left ( 4n^{2} + 6n – 1 \right )}{3}\\$

Now, for n = 1:

1(4.12+6.11)3$\\\frac{1\left ( 4.1^{2} + 6.1 – 1 \right )}{3}$= 4+613$\frac{4 + 6 – 1}{3}$= 93$\frac{9}{3}$ = 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++(2k1)(2k+1)=k(4k2+6k1)3$1.3 + 3.5 + 5.7 + … + \left ( 2k – 1 \right )\left ( 2k + 1 \right ) = \frac{k\left ( 4k^{2} + 6k – 1 \right )}{3}$ . . . . . . . . .  (1)

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

=1.3+3.5+5.7++(2k1)(2k+1)+{2(k+1)1}{2(k+1)+1}$\\1.3 + 3.5 + 5.7 + … + \left ( 2k – 1 \right )\left ( 2k + 1 \right ) + \left \{ 2\left ( k + 1 \right ) – 1 \right \}\left \{ 2\left ( k + 1 \right ) + 1\right \}\\$

=k(4k2+6k1)3+(2k+21)(2k+2+1)$\\\frac{k\left ( 4k^{2} + 6k – 1\right )}{3} + \left ( 2k + 2 – 1 \right )\left ( 2k + 2 + 1 \right )\\$

Now, by using equation (1):

=k(4k2+6k1)3+(2k+1)(2k+3)$\\\frac{k\left ( 4k^{2} + 6k – 1\right )}{3} + \left ( 2k + 1 \right )\left ( 2k + 3 \right )\\$

=k(4k2+6k1)3+(4k2+8k+3)$\\\frac{k\left ( 4k^{2} + 6k – 1\right )}{3} + \left ( 4k^{2} + 8k + 3 \right )\\$

=k(4k2+6k1)+3(4k2+8k+3)3$\\\frac{k\left ( 4k^{2} + 6k – 1\right )+ 3\left ( 4k^{2} + 8k + 3 \right )}{3}\\$

=4k3+6k2k+12k2+24k+93$\\\frac{4k^{3} + 6k^{2} – k + 12k^{2} + 24k + 9}{3}\\$

=4k3+18k2+23k+93$\\\frac{4k^{3} + 18k^{2} + 23k + 9}{3}\\$

=4k3+14k2+9k+4k2+14k+93$\\\frac{4k^{3} + 14k^{2} + 9k + 4k^{2} + 14k + 9}{3}\\$

=k(4k2+14k+9)+1(4k2+14k+9)3$\\\frac{k\left (4k^{2} + 14k + 9 \right )+ 1\left (4k^{2} + 14k + 9 \right )}{3}\\$

=(k+1)(4k2+14k+9)3$\\\frac{\left (k + 1 \right )\left (4k^{2} + 14k + 9 \right )}{3}\\$

=(k+1)(4k2+8k+4+6k+61)3$\\\frac{\left (k + 1 \right )\left (4k^{2} + 8k + 4 + 6k + 6 – 1 \right )}{3}\\$

=(k+1){4(k2+2k+1)+6(k+1)1}3$\\\frac{\left ( k + 1 \right )\left \{ 4\left ( k^{2} + 2k + 1\right ) + 6\left ( k + 1 \right ) – 1\right \}}{3}\\$

=(k+1){4(k+1)2+6(k+1)1}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+2.22+3.22++n.2n=(n1)2n+1+2$1.2 + 2.2^{2} + 3.2^{2} + … + n.2^{n} = \left ( n – 1 \right )2^{n + 1} + 2$

Sol:

The given statement is:

P(n): 1.2+2.22+3.22++n.2n=(n1)2n+1+2$1.2 + 2.2^{2} + 3.2^{2} + … + n.2^{n} = \left ( n – 1 \right )2^{n + 1} + 2$

Now, for n = 1:

=(11)21+1+2$\left ( 1 – 1 \right )2^{1 + 1} + 2$ = 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+2.22+3.22++k.2k=(k1)2k+1+2$\\1.2 + 2.2^{2} + 3.2^{2} + … + k.2^{k} = \left ( k – 1 \right )2^{k + 1} + 2$ . . . . . . . . (1)

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

=[1.2+2.22+3.22++k.2k]+(k+1).2k+1$[ 1.2 + 2.2^{2} + 3.2^{2} + … + k.2^{k} ] + ( k + 1 ). 2^{k + 1}\\$

=(k1)2k+1+2+(k+1).2k+1$\\\left ( k – 1 \right )2^{k + 1} + 2 + \left ( k + 1 \right ).2^{k + 1}\\$

=2k+1{(k1)+(k+1)}+2$\\2^{k + 1}\left \{ \left ( k – 1 \right ) + \left ( k + 1 \right ) \right \} + 2\\$

=2k+1.2k+2$\\2^{k + 1}.2k + 2\\$

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

={(k+1)1}2(k+1)+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: 12+14+18++12n=112n$\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{n}} = 1 – \frac{1}{2^{n}}$

Sol:

The given statement is:

12+14+18++12n=112n$\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{n}} = 1 – \frac{1}{2^{n}}\\$

Now, for n = 1:

=1121$1 – \frac{1}{2^{1}}$= 12$\frac{1}{2}$

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

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

12+14+18++12k=112k$\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{k}} = 1 – \frac{1}{2^{k}}$ . . . . . . (1)

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

=(12+14+18++12k)+12k+1$\\\left (\frac{1}{2} + \frac{1}{4} +\frac{1}{8} + … + \frac{1}{2^{k}} \right ) + \frac{1}{2^{k + 1}}\\$

=(112k)+12k+1$\\\left (1 – \frac{1}{2^{k}} \right ) + \frac{1}{2^{k + 1}}\\$

Now, by using equation (1):

=112k+121.2k$\\1 – \frac{1}{2^{k}} + \frac{1}{2^{1}.2^{k}}\\$

=112k(112)$\\1 – \frac{1}{2^{k}}\left (1 – \frac{1}{2} \right )\\$

=112k(12)$\\1 – \frac{1}{2^{k}}\left ( \frac{1}{2} \right )\\$

=112k+1$\\1 – \frac{1}{2^{k + 1}}\\$

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.

125+158+1811++1(3n1)(3n+2)=n(6n+4)$\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}$

Solution:

The given Statement is:
Q(n): 125+158+1811++1(3n1)(3n+2)=n(6n+4)$\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}$

Now, for n = 1

Q(1)=125=1(6(1)+4)=125$\\Q (1) = \frac{1}{2\cdot 5} = \frac{1}{(6(1)+4)} = \frac{1}{2\cdot 5}\\$

Thus, Q(1) proves to be true.

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

Q(p):

= 125+158+1811++1(3p1)(3p+2)=p(6p+4)$\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3p-1)(3p+2)} = \frac{p}{(6p+4)}$ . . . . . . . . . . . . . . (1)

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

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

Q (p+1):

= 125+158+1811++1(3p1)(3p+2)+1[3(p+1)1][3(p+1)+2]$\\\frac{1}{2\cdot 5} + \frac{1}{5\cdot 8} + \frac{1}{8\cdot 11} +…… + \frac{1}{(3p-1)(3p+2)} + \frac{1}{[3(p + 1)-1][3(p + 1)+2]}\\$

Now, by using equation (1):

=p(6p+4)+1[3p+31][3p+3+2]$\\\frac{p}{(6p + 4)} + \frac{1}{[3p + 3 – 1]\;[3p + 3 + 2]}\\$

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

=p2(3p+2)+1[3p+31][3p+3+2]$\\\frac{p}{2(3p + 2)} + \frac{1}{[3p + 3 – 1]\;[3p + 3 + 2]}\\$

=p2(3p+2)+1[3p+31][3p+3+2]$\\\frac{p}{2(3p + 2)} + \frac{1}{[3p + 3 – 1]\;[3p + 3 + 2]}\\$

=12(3p+2)[p2+1(3p+5)]$\\\frac{1}{2(3p + 2)}\left [ \frac{p}{2} +\frac{1}{(3p + 5)} \right ]\\$

=12(3p+2)[p(3p+5)+2(3p+5)]$\\\frac{1}{2(3p + 2)}\left [\frac{p(3p + 5) + 2}{(3p + 5)} \right ]\\$

=12(3p+2)[3p2+5p+2(3p+5)]$\\\frac{1}{2(3p + 2)}\left [\frac{3p^{2} + 5p + 2}{(3p + 5)} \right ]\\$

=12(3p+2)[(3p+2)(p+1)(3p+5)]$\\\frac{1}{2(3p + 2)}\left [\frac{(3p + 2)(p + 1)}{(3p + 5)} \right ]\\$

=p+16p+10$\\\frac{p + 1}{6p + 10}\\$

=p+16(p+1)+4$\\\frac{p + 1}{6(p + 1) + 4}\\$

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.

1123+1234+1345+.+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot 4\cdot 5} + …….+ \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$

Solution:

The given statement is:

Q(n): 1123+1234+1345+.+1n(n+1)(n+2)=n(n+3)4(n+1)(n+2)$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot 4\cdot 5} + …….+ \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$

Now, for n = 1:

Q(1)=1123=(1)((1)+3)4((1)+1)((1)+2)=1123$Q(1) = \frac{1}{1\cdot 2\cdot 3} = \frac{(1)((1)+3)}{4((1)+1)((1)+2)} = \frac{1}{1\cdot 2\cdot 3}\\$

Thus, Q(1) proves to be true.

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

Q (p):

=1123+1234+1345+.+1p(p+1)(p+2)=p(p+3)4(p+1)(p+2)$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \frac{1}{3\cdot \cdot 4\cdot 5} + …….+ \frac{1}{p(p + 1)(p + 2)} = \frac{p(p + 3)}{4(p + 1)(p+2)}$. . . . . . . . . . . (1)

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

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

Q (p + 1):

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

Now, using equation (1):

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

=1(p+1)(p+2)[p(p+3)2+44(p+3)]$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p + 3)^{2} + 4}{4(p + 3)} \right ]\\$

=1(p+1)(p+2)[p(p2+6p+9)+44(p+3)]$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p^{2} + 6p + 9) + 4}{4(p + 3)} \right ]\\$

=1(p+1)(p+2)[p(p2+6p+9)+44(p+3)]$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p^{2} + 6p + 9) + 4}{4(p + 3)} \right ]\\$

=1(p+1)(p+2)[p3+6p2+9p+44(p+3)]$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p^{3} + 6p^{2} + 9p + 4}{4(p + 3)} \right ]\\$

=1(p+1)(p+2)[p(p2+2p+1)+4(p2+2p+1)4(p+3)]$\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p^{2} + 2p + 1) + 4(p^{2} + 2p + 1)}{4(p + 3)} \right ]\\$

=1(p+1)(p+2)[p(p+1)2+4(p+1)24(p+3)]$\\\frac{1}{(p + 1)(p + 2)}\left [ \frac{p(p + 1)^{2} + 4(p + 1)^{2}}{4(p + 3)} \right ]\\$

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

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

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+ar2+.+arn1=a(rn1)r1$a + ar + ar^{2} + ……. + ar^{n-1} = \frac{a(r^{n}-1)}{r-1}$

Solution:

The given statement is:

Q(n): a+ar+ar2+.+arn1=a(rn1)r1$a + ar + ar^{2} + ……. + ar^{n-1} = \frac{a(r^{n}-1)}{r-1}$

Now, for n = 1

Q(1)=a=a(r(1)1)r1=a$Q(1) = a = \frac{a(r^{(1)}-1)}{r-1} = a\\$

Thus, Q (1) proves to be true.

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

Q (p)  = a+ar+ar2+.+arp1=a(rp1)r1$a + ar + ar^{2} + ……. + ar^{p-1} = \frac{a(r^{p}-1)}{r-1}$ . . . . . . . . (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+ar2+.+arp1+ar(p+1)1$a + ar + ar^{2} + ……. + ar^{p-1} + ar^{(p + 1)-1}$

Now, using Equation (1):

=a(rp1)r1+arp$\\\frac{a(r^{p} – 1)}{r – 1} + ar^{p}\\$

=a(rp1)+arp(r1)r1$\\\frac{a(r^{p} – 1) + ar^{p}(r – 1)}{r – 1}\\$

=a(rp1)