Ncert Solutions Maths class 11th

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

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.

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

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

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)= $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+?+\frac{1}{(3k-1)(3k+2)}=\frac{k}{(6k+4)}$ (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.

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

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

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.

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

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

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

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.

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

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.

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 numbers n.