Prove the following by the principle of mathematical induction:

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

Now substituting the from equation i)

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

Now substituting the from equation i)

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

Now substituting the from equation i)

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

Now substituting the from equation i)

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1 , \text{ L.H.S } = 1 \text{ R.H.S } = 1$

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

Now substituting the from equation i)

$\displaystyle = k^2 + 2k+1$

$\displaystyle =(k+1)^2$

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1 , \text{ L.H.S } = 3 \text{ R.H.S } = 3$

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1 , \text{ L.H.S } = 3 \text{ R.H.S } = 3$

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1 , \text{ L.H.S } = 1 \text{ R.H.S } = 1$

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

$\displaystyle \therefore \text{ L.H.S} = \text{ R.H.S} P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1 , \text{ L.H.S } = 25-1 = 24$ which is divisible by 24

Hence $\displaystyle P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \$

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

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

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

Hence $\displaystyle P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

Hence $\displaystyle P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

Hence $\displaystyle P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

$\displaystyle = 72 \lambda + 64 k + 64$

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

$\displaystyle \text{For } n = 1 , \text{ L.H.S } = ab$ RHS $\displaystyle = ab$ Therefore LHS = RHS

Hence $\displaystyle P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \\$

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

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

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

Hence $\displaystyle P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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

$\displaystyle \$

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

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

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

Hence $\displaystyle P(1) \text{ is true for } n = 1$

$\displaystyle \text{Let } P(n) \text{ is true for } n = k$

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

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

$\displaystyle \text{Now we have to show that} P(n) \text{ is true for } n = k+1$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow P(n) \text{ is true for } n = k+1$

Hence by the principle of mathematical induction

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