Prove the following by the principle of mathematical induction:

Question 1: $1 + 2 + 3 + \ldots + n =$ $\frac{n(n+1)}{2}$ i.e., the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

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

For $n = 1$, L.H.S $= 1$ R.H.S $=$ $\frac{1(1+1)}{2}$ $= 1$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $1 + 2 + 3 + \ldots + k =$ $\frac{k(k+1)}{2}$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$1 + 2 + 3 + \ldots + k +(k+1)$

Now substituting the from equation i)

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

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

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

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1 + 2 + 3 + \ldots + n =$ $\frac{n(n+1)}{2}$ is true for all $n \in N$

$\\$

Question 2: $1^2 + 2^2 + 3^2 + \ldots + n^2 =$ $\frac{n(n+1)(2n+1)}{6}$

Let $P(n) :$ $1^2 + 2^2 + 3^2 + \ldots + n^2 =$ $\frac{n(n+1)(2n+1)}{6}$

For $n = 1$, L.H.S $= 1$ R.H.S $=$ $\frac{1(1+1)(2+1)}{6}$ $= 1$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $1^2 + 2^2 + 3^2 + \ldots + k^2 =$ $\frac{k(k+1)(2k+1)}{6}$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$1^2 + 2^2 + 3^2 + \ldots + k^2 + (k+1)^2$

Now substituting the from equation i)

$=$ $\frac{k(k+1)(2k+1)}{6}$ $+ (k+1)^2$

$=$ $\frac{k+1}{6}$ $[ k(2k+1) + 6(k+1) ]$

$=$ $\frac{k+1}{6}$ $[ 2k^2 + k + 6k + 6]$

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

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1^2 + 2^2 + 3^2 + \ldots + n^2 =$ $\frac{n(n+1)(2n+1)}{6}$ is true for all $n \in N$

$\\$

Question 3: $1+ 3 + 3^2 + \ldots + 3^{n-1} =$ $\frac{3^n -1}{2}$

Let $P(n) :$ $1+ 3 + 3^2 + \ldots + 3^{n-1} =$ $\frac{3^n -1}{2}$

For $n = 1$, L.H.S $= 1$ R.H.S $=$ $\frac{3-1}{2}$ $= 1$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $1+ 3 + 3^2 + \ldots + 3^{k-1} =$ $\frac{3^k -1}{2}$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

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

Now substituting the from equation i)

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

$=$ $\frac{3^k - 1 + 2.3^k}{2}$

$=$ $\frac{3.3^k - 1}{2}$

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1+ 3 + 3^2 + \ldots + 3^{n-1} =$ $\frac{3^n -1}{2}$ is true for all $n \in N$

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Question 4: $\frac{1}{1.2}$ $+$ $\frac{1}{2.3}$ $+$ $\frac{1}{3.4}$ $+ \ldots +$ $\frac{1}{n(n+1)}$ $=$ $\frac{n}{n+1}$

Let $P(n) :$ $\frac{1}{1.2}$ $+$ $\frac{1}{2.3}$ $+$ $\frac{1}{3.4}$ $+ \ldots +$ $\frac{1}{n(n+1)}$ $=$ $\frac{n}{n+1}$

For $n = 1$, L.H.S $=$ $\frac{1}{1.2}$ R.H.S $=$ $\frac{1}{2}$ $= 1$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $\frac{1}{1.2}$ $+$ $\frac{1}{2.3}$ $+$ $\frac{1}{3.4}$ $+ \ldots +$ $\frac{1}{k(k+1)}$ $=$ $\frac{k}{k+1}$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow$ $\frac{1}{1.2}$ $+$ $\frac{1}{2.3}$ $+$ $\frac{1}{3.4}$ $+ \ldots +$ $\frac{1}{k(k+1)}$ $+$ $\frac{1}{(k+1)(k+2)}$

Now substituting the from equation i)

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

$=$ $\frac{k}{k+1}$ $\Big[ k+$ $\frac{1}{k+2}$ $\Big]$

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

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

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

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{1}{1.2}$ $+$ $\frac{1}{2.3}$ $+$ $\frac{1}{3.4}$ $+ \ldots +$ $\frac{1}{n(n+1)}$ $=$ $\frac{n}{n+1}$ is true for all $n \in N$

$\\$

Question 5: $1+ 3+ 5 + \ldots + (2n-1) = n^2$ i.e., the sum of firs $n$ odd numbers is $n^2$

Let $P(n) :$ $1+ 3+ 5 + \ldots + (2n-1) = n^2$

For $n = 1$, L.H.S $= 1$ R.H.S $= 1$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $1+ 3+ 5 + \ldots + (2k-1) = k^2$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$1+ 3+ 5 + \ldots + (2k-1) + (2k+1)$

Now substituting the from equation i)

$= k^2 + 2k+1$

$=(k+1)^2$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1+ 3+ 5 + \ldots + (2n-1) = n^2$ is true for all $n \in N$

$\\$

Question 6: $\frac{1}{2.5}$ $+$ $\frac{1}{5.8}$ $+$ $\frac{1}{8.11}$ $+$ $\ldots +$ $\frac{1}{(3n-1)(3n+2)}$ $=$ $\frac{n}{6n+4}$

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

For $n = 1$, L.H.S $=$ $\frac{1}{10}$ R.H.S $=$ $\frac{1}{10}$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $\frac{1}{2.5}$ $+$ $\frac{1}{5.8}$ $+$ $\frac{1}{8.11}$ $+$ $\ldots +$ $\frac{1}{(3k-1)(3k+2)}$ $=$ $\frac{k}{6k+4}$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\frac{1}{2.5}$ $+$ $\frac{1}{5.8}$ $+$ $\frac{1}{8.11}$ $+$ $\ldots +$ $\frac{1}{(3k-1)(3k+2)}$ $+$ $\frac{1}{(3k+2)(3k+5)}$

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

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

$=$ $\frac{1}{3k+2}$ $\Big[$ $\frac{k}{2}$ $+$ $\frac{1}{3k+5}$ $\Big]$

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

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

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

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{1}{2.5}$ $+$ $\frac{1}{5.8}$ $+$ $\frac{1}{8.11}$ $+$ $\ldots +$ $\frac{1}{(3n-1)(3n+2)}$ $=$ $\frac{n}{6n+4}$ is true for all $n \in N$

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Question 7: $\frac{1}{1.4}$ $+$ $\frac{1}{4.7}$ $+$ $\frac{1}{7.10}$ $+ \ldots +$ $\frac{1}{(3n-2)(3n+1)}$ $=$ $\frac{n}{3n+1}$

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

For $n = 1$, L.H.S $=$ $\frac{1}{4}$ R.H.S $=$ $\frac{1}{4}$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

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

Now we have to show that $P(n)$ is true for $n = k+1$

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

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

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

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

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

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

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{1}{1.4}$ $+$ $\frac{1}{4.7}$ $+$ $\frac{1}{7.10}$ $+ \ldots +$ $\frac{1}{(3n-2)(3n+1)}$ $=$ $\frac{n}{3n+1}$ is true for all $n \in N$

$\\$

Question 8: $\frac{1}{3.5}$ $+$ $\frac{1}{5.7}$ $+$ $\frac{1}{7.9}$ $+ \ldots +$ $\frac{1}{(2n+1)(2n+3)}$ $=$ $\frac{n}{3(2n+3)}$

Let $P(n) :$ $\frac{1}{3.5}$ $+$ $\frac{1}{5.7}$ $+$ $\frac{1}{7.9}$ $+ \ldots +$ $\frac{1}{(2n+1)(2n+3)}$ $=$ $\frac{n}{3(2n+3)}$

For $n = 1$, L.H.S $=$ $\frac{1}{15}$ R.H.S $=$ $\frac{1}{15}$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\frac{1}{3.5}$ $+$ $\frac{1}{5.7}$ $+$ $\frac{1}{7.9}$ $+ \ldots +$ $\frac{1}{(2k+1)(2k+3)}$ $=$ $\frac{k}{3(2k+3)}$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\\$

Question 9: $\frac{1}{3.7}$ $+$ $\frac{1}{7.11}$ $+$ $\frac{1}{11.15}$ $+ \ldots +$ $\frac{1}{(4n-1)(4n+3)}$ $=$ $\frac{n}{3(4n+3)}$

Let $P(n) :$ $\frac{1}{3.7}$ $+$ $\frac{1}{7.11}$ $+$ $\frac{1}{11.15}$ $+ \ldots +$ $\frac{1}{(4n-1)(4n+3)}$ $=$ $\frac{n}{3(4n+3)}$

For $n = 1$, L.H.S $=$ $\frac{1}{21}$ R.H.S $=$ $\frac{1}{21}$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $\frac{1}{3.7}$ $+$ $\frac{1}{7.11}$ $+$ $\frac{1}{11.15}$ $+ \ldots +$ $\frac{1}{(4k-1)(4k+3)}$ $=$ $\frac{k}{3(4k+3)}$… … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\frac{1}{3.7}$ $+$ $\frac{1}{7.11}$ $+$ $\frac{1}{11.15}$ $+ \ldots +$ $\frac{1}{(4k-1)(4k+3)}$ $+$ $\frac{1}{[4(k+1)-1][4(k+1)+3]}$

$=$ $\frac{k}{3(4k+3)}$ $+$ $\frac{1}{(4k+3)(4k+7)}$

$=$ $\frac{1}{(4k+3)}$ $\Big[$ $\frac{k}{3}$ $+$ $\frac{1}{4k+7}$ $\Big]$

$=$ $\frac{1}{(4k+3)}$ $\Big[$ $\frac{4k^2+7k+3}{3(4k+7)}$ $\Big]$

$=$ $\frac{1}{(4k+3)}$ $\Big[$ $\frac{(4k+3)(k+1)}{3(4k+7)}$ $\Big]$

$=$ $\frac{k+1}{3(4k+7)}$

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{1}{3.7}$ $+$ $\frac{1}{7.11}$ $+$ $\frac{1}{11.15}$ $+ \ldots +$ $\frac{1}{(4n-1)(4n+3)}$ $=$ $\frac{n}{3(4n+3)}$ is true for all $n \in N$

$\\$

Question 10: $1.2 + 2.2^2 + 3.2^3 + \ldots + n.2^n = ( n-1)2^{n+1} + 2$

Let $P(n) :$ $1.2 + 2.2^2 + 3.2^3 + \ldots + n.2^n = ( n-1)2^{n+1} + 2$

For $n = 1$, L.H.S $= 2$ R.H.S $=$ $2$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $1.2 + 2.2^2 + 3.2^3 + \ldots + k.2^k = ( k-1)2^{k+1} + 2$… … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$1.2 + 2.2^2 + 3.2^3 + \ldots + k.2^k + (k+1)2^{k+1}$

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

$= k2^{k+1}- 2^{k+1}+2 + k 2^{k+1}+ 2^{k+1}$

$= 2 k 2^{k+1}+ 2$

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1.2 + 2.2^2 + 3.2^3 + \ldots + n.2^n = ( n-1)2^{n+1} + 2$ is true for all $n \in N$

$\\$

Question 11: $2 + 5 + 8 + 11 + \ldots + (3n-1) =$ $\frac{1}{2}$ $n(3n+1)$

Let $P(n) :$ $2 + 5 + 8 + 11 + \ldots + (3n-1) =$ $\frac{1}{2}$ $n(3n+1)$

For $n = 1$, L.H.S $= 2$ R.H.S $=$ $2$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $2 + 5 + 8 + 11 + \ldots + (3k-1) =$ $\frac{1}{2}$ $k(3k+1)$

… … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$2 + 5 + 8 + 11 + \ldots + (3k-1) + [3(k+1)-1]$

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

$=$ $\frac{3k^2+7k+4}{2}$

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

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$2 + 5 + 8 + 11 + \ldots + (3n-1) =$ $\frac{1}{2}$ $n(3n+1)$ is true for all $n \in N$

$\\$

Question 12: $1.3 + 2.4 + 3.5 + \ldots + n(n+2) =$ $\frac{1}{6}$ $n(n+1)(2n+7)$

Let $P(n) :$ $1.3 + 2.4 + 3.5 + \ldots + n(n+2) =$ $\frac{1}{6}$ $n(n+1)(2n+7)$

For $n = 1$, L.H.S $= 3$ R.H.S $=$ $3$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $1.3 + 2.4 + 3.5 + \ldots + k(k+2) =$ $\frac{1}{6}$ $k(k+1)(2k+7)$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$1.3 + 2.4 + 3.5 + \ldots + k(k+2) + ( k+1)(k+3)$

$=$ $\frac{1}{6}$ $k(k+1)(2k+7) + ( k+1)(k+3)$

$= (k+1) \Big[$ $\frac{k}{6}$ $(2k+7)+(k+3) \Big]$

$= (k+1) \Big[$ $\frac{2k^2+13k+18}{6}$ $\Big]$

$=$ $\frac{1}{6}$ $(k+1) (2k+9)(k+2)$

$=$ $\frac{1}{6}$ $(k+1) [( k+1)+1] [2(k+1)+7]$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1.3 + 2.4 + 3.5 + \ldots + n(n+2) =$ $\frac{1}{6}$ $n(n+1)(2n+7)$ is true for all $n \in N$

$\\$

Question 13: $1.3 + 3.5 + 5.7 + \ldots + (2n-1)(2n+1) =$ $\frac{n(4n^2+6n-1)}{3}$

Let $P(n) :$ $1.3 + 3.5 + 5.7 + \ldots + (2n-1)(2n+1) =$ $\frac{n(4n^2+6n-1)}{3}$

For $n = 1$, L.H.S $= 3$ R.H.S $= 3$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $1.3 + 3.5 + 5.7 + \ldots + (2k-1)(2k+1) =$ $\frac{k(4k^2+6k-1)}{3}$

… … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$1.3 + 3.5 + 5.7 + \ldots + (2k-1)(2k+1) + [2(k+1)-1][2(k+1)+1]$

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

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

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

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

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1.3 + 3.5 + 5.7 + \ldots + (2n-1)(2n+1) =$ $\frac{n(4n^2+6n-1)}{3}$ is true for all $n \in N$

$\\$

Question 14: $1.2 + 2.3 + 3.4 + \ldots + n(n+1) =$ $\frac{n(n+1)(n+2)}{3}$

Let $P(n) :$ $1.2 + 2.3 + 3.4 + \ldots + n(n+1) =$ $\frac{n(n+1)(n+2)}{3}$

For $n = 1$, L.H.S $= 2$ R.H.S $= 2$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $1.2 + 2.3 + 3.4 + \ldots + k(k+1) =$ $\frac{k(k+1)(k+2)}{3}$

… … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$1.2 + 2.3 + 3.4 + \ldots + k(k+1) + (k+1)(k+2)$

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

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

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

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1.2 + 2.3 + 3.4 + \ldots + n(n+1) =$ $\frac{n(n+1)(n+2)}{3}$ is true for all $n \in N$

$\\$

Question 15: $\frac{1}{2}$ $+$ $\frac{1}{4}$ $+$ $\frac{1}{8}$ $+ \ldots +$ $\frac{1}{2^n}$ $= 1 -$ $\frac{1}{2^n}$

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

For $n = 1$, L.H.S $=$ $\frac{1}{2}$ R.H.S $=$ $\frac{1}{2}$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $\frac{1}{2}$ $+$ $\frac{1}{4}$ $+$ $\frac{1}{8}$ $+ \ldots +$ $\frac{1}{2^k}$ $= 1 -$ $\frac{1}{2^k}$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\frac{1}{2}$ $+$ $\frac{1}{4}$ $+$ $\frac{1}{8}$ $+ \ldots +$ $\frac{1}{2^k}$ $+$ $\frac{1}{2^{k+1}}$

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

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

$= 1 -$ $\frac{1}{2^{k+1}}$$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$\frac{1}{2}$ $+$ $\frac{1}{4}$ $+$ $\frac{1}{8}$ $+ \ldots +$ $\frac{1}{2^n}$ $= 1 -$ $\frac{1}{2^n}$ is true for all $n \in N$

$\\$

Question 16: $1^2 + 3^2 + 5^2 + \ldots +(2 n-1)^2 =$ $\frac{1}{3}$ $n(4n^2-1)$

Let $P(n) :$ $1^2 + 3^2 + 5^2 + \ldots +(2 n-1)^2 =$ $\frac{1}{3}$ $n(4n^2-1)$

For $n = 1$, L.H.S $= 1$ R.H.S $= 1$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$\Rightarrow$ $1^2 + 3^2 + 5^2 + \ldots +(2 k-1)^2 =$ $\frac{1}{3}$ $k(4k^2-1)$

… … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$1^2 + 3^2 + 5^2 + \ldots +(2 n-1)^2 + [ 2(k+1)-1]^2$

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

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

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

$= (2k+1) \big[$ $\frac{2k^2 - k + 6k + 3}{3}$ $\Big]$

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

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

$=$ $\frac{k+1}{3}$ $[ 4k^2 + 8k + 3]$

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$1^2 + 3^2 + 5^2 + \ldots +(2 n-1)^2 =$ $\frac{1}{3}$ $n(4n^2-1)$ is true for all $n \in N$

$\\$

Question 17: $a+ar+ar^2+ \ldots + ar^{n-1} = a \Big($ $\frac{r^n - 1}{r-1}$ $\Big) , r \neq 1$

Let $P(n) :$ $a+ar+ar^2+ \ldots + ar^{n-1} = a \Big($ $\frac{r^n - 1}{r-1}$ $\Big) , r \neq 1$

For $n = 1$, L.H.S $= a$ R.H.S $= a \Big($ $\frac{r^1 - 1}{r-1}$ $\Big) = a$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$a+ar+ar^2+ \ldots + ar^{k-1} = a \Big($ $\frac{r^k - 1}{r-1}$ $\Big) , r \neq 1$… … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$a+ar+ar^2+ \ldots + ar^{n-1} + ar^k$

$= a \Big($ $\frac{r^k-1}{r-1}$ $\Big) + ar^k$

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

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

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$a+ar+ar^2+ \ldots + ar^{n-1} = a \Big($ $\frac{r^n - 1}{r-1}$ $\Big) , r \neq 1$ is true for all $n \in N$

$\\$

Question 18: $a+ (a+d) + ( a+ 2d) + \ldots + (a+(n-1)d) =$ $\frac{n}{2}$ $[2a+ (n-1)d]$

Let $P(n) :$ $a+ (a+d) + ( a+ 2d) + \ldots + (a+(n-1)d) =$ $\frac{n}{2}$ $[2a+ (n-1)d]$

For $n = 1$, L.H.S $= a$ R.H.S $=$ $\frac{1}{2}$ $[2a+(1-1)d] = a$

$\therefore$ L.H.S $=$ R.H.S Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$a+ (a+d) + ( a+ 2d) + \ldots + (a+(k-1)d) =$ $\frac{k}{2}$ $[2a+ (k-1)d]$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$a+ (a+d) + ( a+ 2d) + \ldots + (a+(k-1)d) + [ a + kd]$

$=$ $\frac{k}{2}$ $[ 2a + (k-1) d] + [a+kd]$

$= ak +$ $\frac{k(k-1)d}{2}$ $+ a + kd$

$= a(k+1) +$ $\frac{k^2d-kd+2kd}{2}$

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

$=$ $\frac{1}{2}$ $(k+1) [ 2a+kd]$

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

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$a+ (a+d) + ( a+ 2d) + \ldots + (a+(n-1)d) =$ $\frac{n}{2}$ $[2a+ (n-1)d]$ is true for all $n \in N$

$\\$

Question 19: $5^{2n} -1$ is divisible by $24$ for all $n \in N$

Let $P(n) :$ $5^{2n} -1$ is divisible by $24$ for all $n \in N$

For $n = 1$, L.H.S $= 25-1 = 24$ which is divisible by 24

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$5^{2k} -1$ is divisible by $24$ for all $k \in N$

$\therefore 5^{2k} -1 = 24 \lambda \Rightarrow 5^{2k} = 24 \lambda +1$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow 5^{2(k+1)} - 1 = 25. 5^{2k} - 1 = 25 (24 \lambda +1) - 1 = 25 \times 24 \lambda + 24 \Rightarrow 24 ( 25 \lambda + 1) 24 \mu$ where $\mu = 25 \lambda + 1$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$5^{2n} -1$ is divisible by $24$ for all $n \in N$

$\$

Question 20: $3^{2n} +7$ is divisible by $8$ for all $n \in N$

Let $P(n) :$ $3^{2n} +7$ is divisible by $8$ for all $n \in N$

For $n = 1$, L.H.S $=3^2+7=16$ which is divisible by $8$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$3^{2k} +7$ is divisible by $8$ for all $k \in N$

$\therefore 3^{2k}+7 = 8 \lambda \Rightarrow 3^{2k} = 8 \lambda -7$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow 3^{2(k+1)} +7 = 9. 3^{2k} +7 = 9 (8 \lambda -7) +7 = 72 \lambda -56 \Rightarrow 8 ( 9 \lambda - 7) = 8 \mu$ where $\mu = 9 \lambda - 7$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$3^{2n} +7$ is divisible by $8$ for all $n \in N$

$\\$

Question 21: $5^{2n+2} -24n - 25$ is divisible by $576$ for all $n \in N$

Let $P(n) :$ $5^{2n+2} -24n - 25$ is divisible by $576$ for all $n \in N$

For $n = 1$, L.H.S $= 5^4 -24 - 25 = 625- 49 = 576$ which is divisible by $576$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$5^{2k+2} -24k - 25$ is divisible by $576$ for all $k \in N$

$\therefore 5^{2k+2} -24k - 25 = 576 \lambda$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow 5^{2(k+1)+2} -24(k+1) - 25$

$= 5^{2k+2} . 5^2 - 24k - 24 -25$

$= 25 . 5^{2k+2} - 24k-49$

$= 25 (576 \lambda + 24k + 25) - 24k - 49$

$= 25.576 \lambda + 600k + 625 - 24 k - 49$

$= 25. 576 \lambda + 576 k + 576$

$= 576 ( 25 \lambda + k + 1)$

$= 576 \mu$ where $\mu = 25 \lambda + k + 1$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$5^{2n+2} -24n - 25$ is divisible by $576$ for all $n \in N$

$\\$

Question 22: $3^{2n+2} -8n - 9$ is divisible by $8$ for all $n \in N$

Let $P(n) :$ $3^{2n+2} -8n - 9$ is divisible by $8$ for all $n \in N$

For $n = 1$, L.H.S $= 3^4 -8 - 9 = 81- 17 = 64$ which is divisible by $8$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$3^{2k+2} -8k - 9$ is divisible by $8$ for all $n \in N$

$\therefore 3^{2k+2} -8k - 9 = 8 \lambda \Rightarrow 3^{2k+2} = 8 \lambda +8k+9$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

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

$= 3^{2k+2} . 3^2 - 8k - 8 -9$

$= 9 . 3^{2k+2} - 8k-17$

$= 9 (8 \lambda + 8k + 9) - 8k - 17$

$= 72 \lambda + 72k + 81 - 8 k - 17$

$= 72 \lambda + 64 k + 64$

$= 8 ( 9 \lambda + 8k + 8)$

$= 8 \mu$ where $\mu = 9 \lambda + 8k + 8$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$3^{2n+2} -8n - 9$ is divisible by $8$ for all $n \in N$

$\\$

Question 23: $(ab)^n = a^n b^n$ for all $n \in N$

Let $P(n) :$ $(ab)^n = a^n b^n$ for all $n \in N$

For $n = 1$, L.H.S $= ab$  RHS $= ab$ Therefore LHS = RHS

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$(ab)^k = a^k b^k$ for all $k \in N$

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow (ab)^{k+1} = (ab)^k (ab) = (a^k b^k) ( ab) = a^{k+1}b^{k+1} = (ab)^{k+1}$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$(ab)^n = a^n b^n$ for all $n \in N$

$\\$

Question 24: $n(n+1)(n+5)$ is a multiple of $3$ for all $n \in N$

Let $P(n) :$ $n(n+1)(n+5)$ is a multiple of $3$ for all $n \in N$

For $n = 1$, L.H.S $= 1 \times 2 \times 6 = 12$ which is divisible by $3$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$k(k+1)(k+5)$ is divisible by $3$ for all $k \in N$

$\therefore k(k+1)(k+5) = 3 \lambda \Rightarrow k^3+6k^2+5k = 3 \lambda$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow (k+1)(k+2)(k+6)$

$= (k^2 +3k+2)(k+6)$

$= k^3+6k^2+3k^2+6k+2k+12$

$= (k^3+6k^2+5k)+(3k^2+3k+12)$

$= 3 \lambda+3(k^2+k+4)$

$= 3 \mu$ where $\mu = \lambda+(k^2+k+4)$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$n(n+1)(n+5)$ is a multiple of $3$ for all $n \in N$

$\$

Question 25: $7^{2n}+2^{3n-3} . 3^{n-1}$ is divisible by $25$ for all $n \in N$

Let $P(n) :$ $7^{2n}+2^{3n-3} . 3^{n-1}$ is divisible by $25$ for all $n \in N$

For $n = 1$, L.H.S $= 49 + 2^0 \times 3^0 = 50$ which is divisible by $25$

Hence $P(1)$ is true for $n = 1$

Let $P(n)$ is true for $n = k$

$7^{2k}+2^{3k-3} . 3^{k-1}$ is divisible by $25$ for all $k \in N$

$\therefore 7^{2k}+2^{3k-3} . 3^{k-1} = 25 \lambda$ … … … … … i)

Now we have to show that $P(n)$ is true for $n = k+1$

$\Rightarrow 7^{2(k+1)}+2^{3(k+1)-3} . 3^{(+1)k-1}$

$= 49 \times 7^{2k} + 2^{3k} 3^k$

$= 49 ( 25 \lambda - 2^{3k-3} . 3^{k-1} ) + 2^{3k} 3^k$

$= 49 \times 25 \lambda - 49 \Big( \frac{2^{2k}}{8} \Big) \Big( \frac{3^k}{3} \Big) + 2^{3k} 3^k$

$= 49 \times 25 \lambda -2^{3k} 3^k ( \frac{49}{24} -1)$

$= 49 \times 25 \lambda -2^{3k} 3^k \Big( \frac{25}{24} \Big)$

$= 25 \mu$ where $\mu = \Big( 49 \lambda - \frac{2^{3k} 3^k}{24} \Big)$

$\Rightarrow P(n)$ is true for $n = k+1$

Hence by the principle of mathematical induction

$7^{2n}+2^{3n-3} . 3^{n-1}$ is divisible by $25$ for all $n \in N$