NCERT Solutions Class 11 Maths Chapter 4 - Principle of Mathematical Induction Solutions

PMI Exercise 4.1 NCERT Solution

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.,

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.

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.

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.

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.

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.

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.

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.

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