NCERT Solutions Class 11 Maths Principle of Mathematical Induction: Questions, Preparation

• Motivation
• Principle of Mathematical Induction

Check the NCERT Solutions Maths Class 11 Chapter 4 Principle of Mathematical Induction below.

Q 4.8 1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1)2^{n + 1} + 2

• A 4.8
• We can write the given statement as
• P(n)=1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹+2
• If n=1, we get
•  P(1) =1.2¹
•  =1.2 = 2 = (1 – 1) 2ⁿ⁺¹+2
•  =2
• which is true.
• Let us assume P(k) is true, for some positive integer k.
• i.e.,1.2 + 2.2² + 3.2² + … + k.2^k = (k – 1) 2^k+1+2-------------------------(1)
• Let us prove that P(k+1) is true,
• 1.2 + 2.2² + 3.2² + … + k.2^k + (k+1) 2^k+1
• By using (1),
• =(k – 1) 2^k+1+2+(k+1) 2^k+1
• =2^k+1{(k – 1)+(k+1)}+2
• =2^k+1.2.k+2
• =k.2^k+1+1+2
• ={(k+1) –1} 2^(k+1)+1+2
• ⸫ P(k+1) is true whenever P(k) is true. Hence, From P.M.I. the P(n) is true for all natural number n.

Q4.19 n(n+1)(n +5)  is a multiple of 3

A 4.19

We can write the given statement as

P(n): n(n +1)(n+5), which is multiple of 3.

If n= 1, we get

P(1)=1(1+1)(1+5)=12, which is a multiple of 3 which is true.

Consider P(k) be true for some positive integer k

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

k(k+1)(k+5)= 3 m, where m ∈ N (1)

Now, let us prove that P(k + 1) is true

Here, 

(k+ 1){(k+1)+ 1}{(k+1)+ 5}

We can write it as

=(k +1)(k+ 2){(k + 5) + 1}

By Multiplying the terms.

(k+1) (k+2) (k+5) + (k+1) (k+2)

{ k(k+1) ( k+5) + 2 (K+1) (K+5) } + (k+1) (k+2)

By eqn. (1)

 = 3m + 2 (k + 1)(k + 5) + (k + 1) (k + 2)

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

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

= 3m + (k + 1) (3k +12)

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

=3{m + (k + 1) (k + 4)}

3  9 where 9 = {m+(k + 1) (k + 4)} is some natural number (k + 1){(k + 1) + 5} is multiple of 3.

P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction, statement P(n) is true for all natural number.

Q 4.24 2n+7<(n  3)²

^{A 4.24}

Let P(n) be the statement “ 2n+7<(n+3)²”

of n=1

P(1): 2 X 1 +7 < (1+3)²

9<4²

9<16 which is true. This P(1) is true.

Suppose P(k) is true.

P(k)= 2k+7<(k+3)²     (1)

Lets prove that P(k +1) is also true.

“ 2(k + 1) + 7 < (k + 4)²=k²+ 8k + 16”

P(k +1) = 2(k +1) +7 = (2k +7) +2

< (k +3)²+ 2                  (Using 1)

= k²+ 9 + 6k +2 = k²+6k +11

Adding and subtracting (2k + k) in the R. H. S.

K²+6k+11+2k+5-(2k-5)

(k²-8k+16)-(2k-5)

(k+4)²-(2k-5)

(K+4)², since 2k+5>0 for all k ∈ N

P (K+1) is true.

By the principle of mathematical induction, P(n) is true for all n ∈ N.

Download Here NCERT Class 11th Maths Chapter 4 Principle of Mathematical Induction Solutions PDF

Exercise. 4.1

Q.1. 1+3+3²+ … +3^{n-1= $\frac{\left(3^n-1\right)}{2}$}

A.1.       Let the given statement be P(n) i.e.,

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

For n=1, P(1)=1= $\frac{\left(3^1-1\right)}{2}=\frac{3-1}{2}=\frac{2}{2}=1$

which is true.

Assume that P(k) is true for some positive integer k i.e.,

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

--------(1)

Now, let us prove that P(k+1) is true.

Here, 1+3+3²+ … +3^k–1+3^(k+1)–1

$\frac{3^k-1}{2}+3^{k+1-1}$

[By using eq (1)]

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

=  $\frac{3^k+2.3^k-1}{2}$

=  $\frac{3^k(1+2)-1}{2}$

=  $\frac{3^k·3-1}{2}=\frac{3^{k+1}-1}{2}$

⸫P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all natural numbers n.

 

Q2.  1³+2³+3³+ … +n³= ${\left(\frac{n(n+1)}{2}\right)}^2$

A.2. Let the given statement be P(n) i.e.,

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

For, n=1, P(n)=1³=1= ${\left(\frac{1(1+1)}{2}\right)}^2={\left(\frac{1.2}{2}\right)}^2=1^2=1$

which is true.

Consider P(k) be true for some positive integer k

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

Now, let us prove that P(k+1) is true.

Here,  1³+2³+3³+ … +k³+(k+1)³

By using eq (1)

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

=  $\frac{k^2(k+1){}^2+4·{(k+1)}^3}{4}$

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

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

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

⸫P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, P(n) is true forall natural numbersn.

 

Q3.  $1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\dots +\frac{1}{(1+2+3+\cdots n)}=\frac{2n}{(n+1)}$

A.3. Let the given statement be P(n) i.e.,

P(n): 1+  $\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\cdots +\frac{1}{(1+2+3+\cdots n)}=\frac{2n}{(n+1)}$

For n=1,

we get,P(1)=1=  $\frac{2.1}{1+1}=\frac{2}{2}=1$

which is true.

Let us assume that P(k) is true for some positive integer k.

i.e.,  $1+\frac{1}{1+2}+\dots +\frac{1}{(1+2+3+\cdots k)}=\frac{2k}{(k+1)}$ ------------------ (1)

Which is true.

Now, let us prove that P(k + 1) is true.

$1+\frac{1}{1+2}+\frac{1}{1+2+3}$ + …  $+\frac{1}{(1+2+3+\cdots k)}+\frac{1}{1+2+3+\cdots +k+(k+1)}$

By using eqn (1)

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

⸫We know that, 1+2+3+ … +n=  $\frac{n(n+1)}{2}$

So, we get

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

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

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

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

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

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

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

⸫ P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all a natural number n.

 

Q4.1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =  $\frac{n(n+1)(n+2)(n+3)}{4}$

A.4. Let the given statement be P(n) i.e.,

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

If n=1, we get

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

which is true.

considerP(k) is true for some positive integer k

1.2.3 + 2.3.4 + … + k(k + 1)(k + 2) =  $\frac{k(k+1)(k+2)(k+3)}{4}$ -------------------(1)

Now, let us prove that P(k+1) is true.

Here,1.2.3 + 2.3.4 + … + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3)

By eqn (1), we get,

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

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

=  $\frac{(k+1)(k+2)(k+3)(k+4)}{4}$

By further Simplification,

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

⸫ P(k+1)is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all natural numbersn.

 

Q5.1.3 + 2.3² + 3.3³ + … + n.3ⁿ = $\frac{(2n-1)3^{n+1}+3}{4}$

A.5. Let the given statement be P(n) i.e.,

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

If n=1, we get

P(1) = 1.3=3=  $\frac{(2.1-1)3^{1+1}+3}{4}$ =  $\frac{1·3^2+3}{4}$ =  $\frac{12}{4}$ =3

which is true.

Consider P(k) be true for some positive integer k

1.3 + 2.3² + 3.3³ + … + k^3k = $\frac{(2k-1)3^{k+1}+3}{4}$ ------------------(1)

Now, let us prove P(k+1) is true.

Here,

1.3 + 2.3² + 3.3³ + … + k^3k + (k + 1)3^{k + 1}

By using eqn. (1)

$\frac{(2k-1)3^{k+1}+3}{4}+(k+1)3^{k+1}$

L.C.M

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

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

=  $\frac{3^{k+1}\{2k-1+4k+4\}+3}{4}$

=  $\frac{3^{k+1}\{6k+3\}+3}{4}$

=  $\frac{3^{k+1}·3(2k+1)+3}{4}$ =  $\frac{3^{k+1+1}\{2k+1\}+3}{4}$ ⸫P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction statement P(n) is true for all natural numbers i.e., n.

 

Q6.1.2 + 2.3 + 3.4 + … + n (n + 1) =  $[\frac{n(n+1)(n+2)}{3}$

A.6. Let the given statement be P(n) i.e.,

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

For n=1,

P(1)=1.2=2=  $\frac{1(1+1)(1+2)}{3}$ =  $\frac{2\times 3}{3}$ =2.

Which is true.

considerP(k) be true for some positive integer k

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

Now, let us prove that P(k+1) is true.

Here, 1.2 + 2.3 + 3.4 + … + k(k + 1) + (k+1)(k+2)

By using (1), we get

=  $\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

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

=  $\frac{(k+1)(k+2)(k+3)}{3}$

By further simplification;  $\frac{(k+1)(k+1+1)(k+1+2)}{3}$

P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction, statement P(n) is true for all natural no. i.e.,n.

 

Q7.1.3+3.5+5.7+ … +(2n – 1)(2n+1)=n  $\frac{\left(4n^2+6n-1\right)}{}$

A.7. Let the given statement be P(n) i.e.,

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

For,n = 1

P(1)=1.3=3=  $\frac{1\left(4.1^2+6.1-1\right)}{3}$ =  $\frac{4+6-1}{3}$ =  $\frac{9}{3}$ =3

Which is true.

Assume that P(k) is true for some positive integer k i.e.,

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

Let us prove that P(k+1) is true,----------------------(1)

1.3 + 3.5 + 5.7 + … + (2k – 1(2k + 1) + [2(k + 1) –1] [2(k + 1) +1]

By (1),

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

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

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

L.C.M.

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

=  $\frac{4k^3+6k^2-k+12k^2+24k+9}{3}$

=  $\frac{4k^3+18k^2+23k+9}{3}$

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

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

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

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

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

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

⸫ P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all natural numbers n.

 

Q8.        1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1)2^{n + 1} + 2

A.8.        We can write the given statement as

              P(n)=1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹+2

              If n=1, we get

              P(1) =1.2¹

              =1.2 = 2 = (1 – 1) 2ⁿ⁺¹+2

              =2

              which is true.

              Let us assume P(k) is true, for some positive integer k.

              i.e.,1.2 + 2.2² + 3.2² + … + k.2^k = (k – 1) 2^k+1+2                        -------------------------(1)

              Let us prove that P(k+1) is true,

              1.2 + 2.2² + 3.2² + … + k.2^k + (k+1) 2^k+1

              By using (1),

              =(k – 1) 2^k+1+2+(k+1) 2^k+1

              =2^k+1{(k – 1)+(k+1)}+2

=2^k+1{k – $\cancel{1}$ +k+  $\cancel{1}$ }+2

=2^k+1.2.k+2

=k.2^k+1+1+2

={(k+1) –1} 2^(k+1)+1+2

⸫ P(k+1) is true whenever P(k) is true. Hence, From P.M.I. the P(n) is true for all natural numbern.

 

Q9.  $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}=1-\frac{1}{2^n}$

A.9. Let the given statement be P(n) l.e.,

P(n)=  $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}=1-\frac{1}{2^n}$

If n=1, we get

P(1)=  $\frac{1}{2}=1-\frac{1}{2^1}=\frac{1}{2}$

which is true.

Consider P(k) be true for some positive integer k.

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^k}=1-\frac{1}{2^k}$ (1)

Now, let us prove that P(k+1) is true.

Here,  $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^k}+\frac{1}{2^{k+1}}$

By using eqn. (1)

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

we can write as,

=  $1-\frac{1}{2^k}+\frac{1}{2^k.2}$

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

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

It can be written as,=  $1-\frac{1}{2^{k+1}}$

P(k + 1) is true whenever P(k) is true.

Hence, From the principle of mathematical induction the P(n) is true for all natural number n.

 

Q10.  $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\dots +\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

A.10. Let the given statement be P(n) i.e.,

$P(n)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots +\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

For n=1,

P(1)=  $\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}=\frac{1}{10}$

which is true.

Assume that P(k) is true for some positive integer k.

i.e.,P(k)=  $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\cdots +\frac{1}{(3k-1)(3k+2)}=\frac{k}{(6k+4)}$ (1)

Now, let us prove P(k+1) is true,

Here,  $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}$ + … +  $\frac{1}{(3k-1)(3k+2)}+\frac{1}{(3(k+1)-1)[3(k+1)+2]}$

By using eqn.(1),

=  $\frac{k}{6k+4}+\frac{1}{(3k+3-1)(3k+3+2)}$

=  $\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}$

Taking 2 as common,

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

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

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

=  $\frac{1}{(3k+2)}\left\{\frac{3k^2+5k+2}{\begin{array}{l}\bcancel{6k}+\bcancel{10}\\ 2(3k+5)\end{array}}\right\}$

=  $\frac{1}{(3k+2)}\left\{\frac{3k^2+3k+2k+2}{2(3k+5)}\right\}$ =  $\frac{1}{(3k+2)}\left\{\frac{3k(k+1)+2(k+1)}{2(3k+5)}\right\}$

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

=  $\frac{(k+1)}{6k+10}$ , so we get

$\frac{(k+1)}{6(k+1)+4}$

P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all natural number.

 

Q11.  $\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … +  $\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

A.11. we can write the given statement as

$P(n)=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … +  $\frac{1}{n(n+1)(n+2)}$ =  $\frac{n(n+3)}{4(n+1)(n+2)}$

If n=1,

P(1)=  $\frac{1}{1.2.3}$ =  $\frac{1}{6}$ =  $\frac{1(1+3)}{4(1+1)(1+2)}$ =  $\frac{4}{4\times 2\times 3}$ =  $\frac{1}{6}$

which is true.

Consider P(k) be true for some positive integer k

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … +  $\frac{1}{k(k+1)(k+2)}$ =  $\frac{k(k+3)}{4(k+1)\left(k+2\right)}$

Let us prove that P(k+1) is true,

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … +  $\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$ .

By equation (1), we get

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

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

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

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

=  $\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+6k^2+9k+4}{4(k+3)}\right\}$

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

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

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

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

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

P(k+1) is true whenever P(k) is true.

Hence, By the principle of mathematical induction, the P(n) is true for all natural number n.

 

Q12. a+ar+ar²+ … +ar^n-1= $\frac{a\left(r^n-1\right)}{r-1}$

A.12. Let the given statement be P(n) i.e.,

P(n)=a+ar+ar²+ … +ar^n-1== $\frac{a\left(r^n-1\right)}{r-1}$

If n = 1, we get

P(1)=a=  $\frac{a\left(r^1-1\right)}{r-1}$ =a

which is true.

Consider P(k) be true for some positive integer k

a+ar+ar²+ … +ar^k-1= $\frac{a\left(r^k-1\right)}{r-1}$ (1)

Now, let us prove that P(k+1) is true.

Here, {a+ar+ar²+ … +ar^k-1}+ar^{(k+1) –1}

By using (1),

=  $\frac{a\left(r^k-1\right)}{r-1}+a{r}^k$

=  $\frac{a\left(r^k-1\right)+a{r}^k(r-1)}{r-1}$

=  $\frac{a{r}^k-a+a{r}^{k+1}-a{r}^k}{r-1}$

=  $\frac{a{r}^{k+1}-a}{r-1}$

=  $\frac{a\left(r^{k+1}-1\right)}{r-1}$

P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e.,

 

Q.13.  $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ …  $\left(1+\frac{(2n+1)}{n^2}\right)={(n+1)}^2$

A.13. We can write given statement as

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

If n=1, we get

P(1):  $\left(1+\frac{3}{1}\right)$ =4=(1+ 1)²=2²=4

which is true.

Consider P(k) be true for some positive integer k.

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ …  $\left(1+\frac{(2k+1)}{k^2}\right)={(k+1)}^2$ (1)

Now, let us prove that P(k+1) is true.

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ …  $\left(1+\frac{(2k+1)}{k^2}\right)+\left(1+\frac{(2(k+1)+1)}{{(k+1)}^2}\right)$

By using (1)

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

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

=(k+1)²+2(k+1)+1

={(k+1)+1}²

P(k+1) is true whenever P(k) is true.

Therefore, by principle of mathematical induction, the P(n) is true for all natural number n.

 

Q14.  $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ …  $\left(1+\frac{1}{n}\right)=(n+1)$

A.14. Let the given statement be P(n) i.e.,

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

If n =1

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

which is true.

Assume that P(k) is true for some positive integer k i.e.,

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

Now, let us prove that P(k+1) is true.

Here,

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

By using (1), we get

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

L.C.M.=(k+1).  $\left(\frac{k+1+1}{k+1}\right)$

= (k+1)+1

⸫ P(k+1) is true whenever P(k) is true.

Therefore from the principle of mathematical induction the P(n) is true for all natural numbers n.

 

Q15. 1²+3²+5²+ … + (2n – 1)²= $\frac{(2n-1)(2n+1)}{3}$

A.15. We can write the given statement as

P(n)=1²+3²+5²+ … + (2n – 1)²= $\frac{n(2n-1)(2n+1)}{3}$

forn=1

P(1)=1²=1= $\frac{1(2.1-1)(2.1+1)}{3}$

=  $\frac{1(1)(3)}{3}=1$ which is true.

Consider P(k) be true for some positive integer k

P(k)=1²+3²+5²+ … + (2n – 1)²= $\frac{k(2k-1)(2k+1)}{3}$ ------------------(1)

Now, let us prove that P(k+1) is true.

Here,

1²+3²+5²+ … +(2k – 1)²+(2(k+1) –1)²

By using (1),

$\frac{k(2k-1)(2k+1)}{3}+{[2k+2-1)}^2$

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

=  $\frac{(2k+1)[k(2k-1)+3(2k+1)]}{3}$

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

=  $\frac{(2k+1)\left(2k^2+5k+3\right)}{3}$

we can write as,

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

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

=  $\frac{(2k+1)(2k+3)(k+1)}{3}$

=  $\frac{(2k+1)(k+1)(2k+3)}{3}$

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

P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all natural number n.

 

Q16.  $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … +  $\frac{1}{(3n-2)(3n+1)}$ =  $\frac{n}{(3n+1)}$

A.16. Let the given statement as

P(n)=  $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … +  $\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$

If n=1, then

P(1)=  $\frac{1}{1.4}$ =  $\frac{1}{4}$ =  $\frac{1}{(3.1+1)}$ =  $\frac{1}{4}$

which is true.

Consider P(k)be true for some positive integer k

P(k)=  $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … +  $\frac{1}{(3k-2)(3k+1)}$ =  $\frac{k}{(3k+1)}$ ------------------(1)

Now, let us prove P(k+1) is true.

P(k+1)=  $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … +  $\frac{1}{(3k-2)(3k+1)}+\frac{1}{[3(k+1)-2][3(k+1)+1]}$

By using (1),

$\frac{k}{(3k+1)}+\frac{1}{(3k+3-2)(3k+3+1)}$

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

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

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

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

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

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

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

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

⸫ P(k+1) is true whenever P(k) is true.

Therefore, from the principle of mathematical induction, the P(n) is true for all natural number n.

 

Q17.  $\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\cdots +\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

A.17. We can write the given statement as:-

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

For n = 1,

We get  $P(1)=\frac{1}{3\cdot 5}=\frac{1}{15}=\frac{1}{3(2\cdot 1+3)}=\frac{1}{3(2+3)}=\frac{1}{3\times 5}=\frac{1}{15}$

Which is true.

Consider P(k) be true for some positive integer k.

$P(K)=\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\dots +\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)}$ (1)

Now, let us prove that P(k+ 1) is true.

Now,

P(k +1) =  $1$  $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}+\frac{1}{[2(k+1)+1][2(k+1)+3]}$

By using (1),

$\begin{array}{l}=\frac{k}{3(2k+3)}+\frac{1}{(2k+2+1)(2k+2+3)}\\ =\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)}\\ =\frac{1}{(2k+3)}\left[\frac{k}{3}+\frac{1}{(2k+5)}\right]\\ =\frac{1}{(2k+3)}\left[\frac{k(2k+5)+3}{3(2k+5)}\right]\end{array}$

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

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

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

$\begin{array}{l}=\frac{1}{(2k+3)}\frac{(2k+3(k+1)}{\cdot (2k+1)\cdot 3}\\ =\frac{k+1}{3(2k+5)}\\ =\frac{(k+1)}{3\{2(k+1)+3\}}\end{array}$

P(k+ 1) is true wheneverP(k) is true.

Therefore, from the principle of mathematical induction, theP(n) is true for all natural number n.

 

Q18.  $1+2+3+\cdots +n<\frac{1}{8}{(2n+1)}^2$

A.18. We can write the given statement as

$p(n):1+2+3+\cdots +n<\frac{1}{8}{(2n+1)}^2$

If n = 1, we get,

$$ P(1): 1 <  $\frac{1}{8}$ (2k + 1)²= 1<  $\frac{1}{8}$ (3)²

= 1 <  $\frac{9}{8}$

Which is true.

Consider P(k) be true some positive integer k

1+ 2 + ….. + k<  $\frac{1}{8}$ (2k + 1)²                                                  (1)

Let us prove P(k +1) is true.

Here,

(1 + 2 +…. k)+ (k +1) <  $\frac{1}{8}$ (2k + 1)²+ (k +1)

By using (1),

$<\frac{1}{8}\left\{{(2k+1)}^2+8(k+1)\right\}$

$<\frac{1}{8}\left\{{(2k)}^2+2\cdot 2k\cdot +1^2+8k+8\right\}$

$<\frac{1}{8}\left\{4k^2+4k+1+8k+8\right\}$

$<\frac{1}{8}\left\{4k^2+12k+9\right\}$

So, we get,

<  $\frac{1}{8}$ {2k+ 3}²

<  $\frac{1}{8}$ {2(k +1) +1}²

(1 + 2 + 3 + … + k) + (k + 1) <  $\frac{1}{8}$ (2k +1)²+ (k +1)

P(k + 1) is true whenever P(k) is true.

$\therefore$ Hence, from the Principle of mathematical induction, the P(k) is true for all natural numbern.

 

Q19. n(n+1)(n +5) is a multiple of 3.

A.19. We can write the given statement as

P(n): n(n +1)(n+5), which is multiple of 3.

If n= 1, we get

P(1)=1(1+1)(1+5)=12, which is a multiple of 3 which is true.

Consider P(k) be true for some positive integer k

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

k(k+1)(k+5)= 3 m, where  $m\in N$ (1)

Now, let us prove that P(k + 1) is true

Here,

(k+ 1){(k+1)+ 1}{(k+1)+ 5}

We can write it as

=(k +1)(k+ 2){(k + 5) + 1}

By Multiplying the terms.

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

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

By eqn. (1)

= 3m + 2 (k + 1)(k + 5) + (k + 1) (k + 2)

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

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

= 3m + (k + 1) (3k +12)

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

=3{m + (k + 1) (k + 4)}

3  $\times$ 9 where 9 = {m+(k + 1) (k + 4)} is some natural number (k + 1){(k + 1) + 5} is multiple of 3.

P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction, statement P(n) is true for all natural number.

 

Q20.  $10^{2n-1}+$ 1 is divisible by 11.

A.20. LetP(n): $10^{2n-1}+$ 1 is divisible by 11.

Putting n = 1

$P(1)=10^{\prime }+1=11$ is divisible by 11.

Which is true. Thus, P(1) is true.

Let us assume that P(k) is true for some natural no. k.

P(k)=  $10^{2k-1}+\mathrm{<br>}\mathrm{<br>}\mathrm{1\; is\; divisible\; by\; 11.}$

$10^{2k-1}+1=11aa\in z$

$\to 10^{2k-1}=11a-1$ (1)

we want to prove that P(k +1) is true.

$P(k+1):10^{2(K+1)-1}+1=10^{2k+1}+1\mathrm{is\; divisible\; by\; 11.}\mathrm{<br>}$

$10^{2k+1}+1=10^{(2k-1)+2}+1$

$=10^{2k-1}\cdot 10^2+1$

$=100(11a-1)+1(using(1))$

=1100a  $-$ 99= 11(100a  $-$ 9)

11bwhere b= (100a  $-$ 9)  $\in z$

$\Rightarrow 1{0^2}^{k+1}+1$ is divisible by 11.

$\Rightarrow P(k+1)\mathrm{<br>}<br>$ is true when p(k) is true.

Hence by P.M.I. P(n) is true for every positive integer.

 

Q21.  $x^{2x}-y^{2x}\mathrm{is\; divisible\; by\; x}+y$

A.21.  Let 

$\mathrm{Putting\; x}=1,$

$P(1)=x^2-y^2\mathrm{is\; divisible\; by\; x}+y$ or

$\left(x+y\right)(x-y)isdivisiblebyx+y,$ which is true.

Assume that P(k) is true for some natural no. k

$P(k)=x^{2k}-y^{2k}$ is divisible by x + y

i.e.  $x^{2k}-y^{2k}=a(x+y)$ where  $\in z$

$\Rightarrow x^{2k}=a(x+y)+y^{2k}$ (1)

Now, let us prove P(k +1) is true.

$P(k+1):x^{2(k+1)}-y^{2(k+1)}$

$=x^{2x+2}-y^{2x+2}$

$=x^2\cdot x^{2k}-y^2\cdot y^{2k}$

$=x^2\left[a(x+y)+y^{2k}\right]-y^2\cdot y^{2k}[using(1)<br>]$

$=a{x}^2(x+y)+x^2\cdot y^{2k}-y^2\cdot y^{2k}$

$=a{x}^2(x+y)+y^{2k}\left(x^2-y^2\right)$

$=a{x}^2(x+y)+y^{2k}(x+y)(x-y)$

$=(x+y)\left[a{x}^2+y^{2k}(x-y)\right]$

$=b(x+y)whereb=\left[a{x}^2+y^{2k}(x-y)\right]\in z$

$\Rightarrow x^{2k+2}-y^{2k+2}\mathrm{is\; divisible\; by\; x}+y$

$\therefore$ P(k+ 1is true where P (k) is true.

Hence, by P.M.I. P(n) is true for all natural number i.e.,

 

Q22.  $3^{2n+2}-8n-9$ is divisible by 8

A.22. Let P(n): $3^{2n+2}-8n-9$ is divisible by 8

put n= 1,

P(1):  $3^{2+2}-8.1-9$

3⁴ – 8 – 9 = 81– 17 = 64= is divisible by 8

Which is true.

Assume that P(k) is true for some natural numbers k.

i.e,  $3^{2k+2}-8k-9$ be divisible by 8

$3^{2k+2}-8k-9=8a$ where,a  $\in z$

$3^{2k+2}=8a+8k+9$ (1)

We want to prove thatP(k+ 1) is true.

$P(k+1):3^{2(k+1)+2}-8(k+1)-9$ is divisible by 8, is also true.

Now,

${3^{2k+2+}}^2-=8(k+1)-9$

=  $3^{2k+4}-8k+8-9$

$3^{(2k+2)+2}-8k+8-9$

3^{(2k +2)}. 3² $-$ 8k  $-$ 17

$=9(8a+8k+9)-8k-17$ (Using 1)

$=72a+72k+81-8k-17$

= 72a + 64k+ 64 = 8(9a + 8k + 8)

= 8b,

Where b = 9a + 8b + 8  $a\in z$

$\Rightarrow$ 3^{2k + 4}– 8(k+1) – 9 is divisible by 8.

$\therefore$ P(k+1) is true when P(k) is true. Hence, By P.M.I. P(n) is true for all positive integer n.

 

Q23.  $41^2-14^2$ is a multiple of 27

A.23. Let  $P(n):41^n-14^n\mathrm{is\; a\; multiple\; of\; 27.}$

Put n= 1,

$P(1)=41-14=27$ is a multiple of 27.

Which is true.

Assume that P(k) is true for some natural no. k.

P(k)=  $41^k-14^k$ be a multiple of 27

i.e,  $41^k-14^k=27a,a\in z$

$\Rightarrow 41^k=27a+14^k$ (1)

We want to prove that P(k+1) is also true.

Now,

$P(k+1)=41^{k+1}-14^{k+1}$

$=41^k\cdot 41-14\cdot 14^k$

$=41\left(27a+14^k\right)-14\cdot 14^k$ (Using 1)

$=41\times 27a+41\cdot 14^k-14\cdot 14^k$

$=-27\left(24+a+14^k\right)=41\times 27a+14^k(41-14)$

$=41\times 27a+27\cdot 14^k$

$=27\left(41a+14^k\right)$

$=27b,whereb=\left(41a+14^k\right)\in z$

$\Rightarrow 41^{k+1}-14^{k+1}\mathrm{is\; multiple\; of\; 2}7$

$\Rightarrow P(k+1)$ is true when P(k) is true.

Hence, by P.M.I. P(n) is true for every positive integer n.

 

Q24.2n+7<(n  3)²

A.24. Let P(n) be the statement “ 2n+7<(n+3)2”

ofn=1

P(1): 2  $\times 1+7<{(1+3)}^2$

$9<4^2$

9<16 which is true. This P(1) is true.

Suppose P(k) is true.

P(k)= 2k+7<(k+3)^{2                                   (1)}

Lets prove that P(k +1) is also true.

“ 2(k + 1) + 7 < (k + 4)²=k²+ 8k + 16”

P(k +1) = 2(k +1) +7 = (2k +7) +2

  < (k +3)²+ 2                  (Using 1)

= k²+ 9 + 6k +2 = k²+6k +11

Adding and subtracting (2k + k) in the R. H. S.

$=k^2+6k+11+2k+5-(2k-5)$

$=\left(k^2-8k+16\right)-(2k-5)$

$={(k+4)}^2-(2k-5)$

$<{(k+4)}^2\mathrm{,since\; 2}k+5>0$ for all $k\in N$

$\Rightarrow P(k+1)$ is true.

$\therefore$ By the principle of mathematical induction, P(n)is true for all n  $\in$ N.