\displaystyle \textbf{Question 1: }\text{Prove by mathematical induction that }1+2+3+\ldots+n=\frac{n(n+1)}{2},
\displaystyle \text{i.e., the sum of the first }n\text{ natural numbers is }\frac{n(n+1)}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1+2+3+\ldots+n=\frac{n(n+1)}{2}
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1
\displaystyle \text{RHS}=\frac{1(1+1)}{2}=1
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1+2+3+\ldots+k=\frac{k(k+1)}{2}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1+2+3+\ldots+k+(k+1)
\displaystyle \text{Using 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 =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1+2+3+\ldots+n=\frac{n(n+1)}{2}\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Prove by mathematical induction that }1^2+2^2+3^2+\ldots+n^2
\displaystyle =\frac{n(n+1)(2n+1)}{6}.
\displaystyle \text{Answer:}
\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,
\displaystyle \text{LHS}=1^2=1
\displaystyle \text{RHS}=\frac{1(1+1)(2\times1+1)}{6}=1
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1^2+2^2+3^2+\ldots+k^2=\frac{k(k+1)(2k+1)}{6}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1^2+2^2+3^2+\ldots+k^2+(k+1)^2
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{k(k+1)(2k+1)}{6}+(k+1)^2
\displaystyle =\frac{k+1}{6}\left[k(2k+1)+6(k+1)\right]
\displaystyle =\frac{k+1}{6}\left(2k^2+7k+6\right)
\displaystyle =\frac{(k+1)(2k+3)(k+2)}{6}
\displaystyle =\frac{(k+1)\{(k+1)+1\}\{2(k+1)+1\}}{6}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{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{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove by mathematical induction that }1+3+3^2+\ldots+3^{n-1}
\displaystyle =\frac{3^n-1}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1+3+3^2+\ldots+3^{n-1}=\frac{3^n-1}{2}
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1
\displaystyle \text{RHS}=\frac{3^1-1}{2}=\frac{3-1}{2}=1
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1+3+3^2+\ldots+3^{k-1}=\frac{3^k-1}{2}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1+3+3^2+\ldots+3^{k-1}+3^k
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{3^k-1}{2}+3^k
\displaystyle =\frac{3^k-1+2\times3^k}{2}
\displaystyle =\frac{3\times3^k-1}{2}
\displaystyle =\frac{3^{k+1}-1}{2}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1+3+3^2+\ldots+3^{n-1}=\frac{3^n-1}{2}\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Prove by mathematical induction that }\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\ldots
\displaystyle +\frac{1}{n(n+1)}=\frac{n}{n+1}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\frac{1}{1\times2}=\frac{1}{2}
\displaystyle \text{RHS}=\frac{1}{1+1}=\frac{1}{2}
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\ldots+\frac{1}{k(k+1)}
\displaystyle =\frac{k}{k+1}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=\frac{1}{1\times2}+\frac{1}{2\times3}+\ldots+\frac{1}{k(k+1)}
\displaystyle +\frac{1}{(k+1)(k+2)}
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}
\displaystyle =\frac{k(k+2)+1}{(k+1)(k+2)}
\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 =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Prove by mathematical induction that }1+3+5+\ldots+(2n-1)=n^2,
\displaystyle \text{i.e., the sum of the first }n\text{ odd natural numbers is }n^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1+3+5+\ldots+(2n-1)=n^2
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1
\displaystyle \text{RHS}=1^2=1
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1+3+5+\ldots+(2k-1)=k^2\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1+3+5+\ldots+(2k-1)+(2k+1)
\displaystyle \text{Using equation (i),}
\displaystyle =k^2+2k+1
\displaystyle =(k+1)^2
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1+3+5+\ldots+(2n-1)=n^2\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Prove by mathematical induction that }\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\ldots
\displaystyle +\frac{1}{(3n-1)(3n+2)}=\frac{n}{6n+4}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\ldots+\frac{1}{(3n-1)(3n+2)}=\frac{n}{6n+4}
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\frac{1}{2\times5}=\frac{1}{10}
\displaystyle \text{RHS}=\frac{1}{6(1)+4}=\frac{1}{10}
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\ldots+\frac{1}{(3k-1)(3k+2)}
\displaystyle =\frac{k}{6k+4}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=\frac{1}{2\times5}+\frac{1}{5\times8}+\ldots+\frac{1}{(3k-1)(3k+2)}
\displaystyle +\frac{1}{(3k+2)(3k+5)}
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}
\displaystyle =\frac{k}{2(3k+2)}+\frac{1}{(3k+2)(3k+5)}
\displaystyle =\frac{1}{3k+2}\left[\frac{k}{2}+\frac{1}{3k+5}\right]
\displaystyle =\frac{1}{3k+2}\left[\frac{k(3k+5)+2}{2(3k+5)}\right]
\displaystyle =\frac{3k^2+5k+2}{2(3k+2)(3k+5)}
\displaystyle =\frac{(3k+2)(k+1)}{2(3k+2)(3k+5)}
\displaystyle =\frac{k+1}{2(3k+5)}
\displaystyle =\frac{k+1}{6k+10}
\displaystyle =\frac{k+1}{6(k+1)+4}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\ldots+\frac{1}{(3n-1)(3n+2)}
\displaystyle =\frac{n}{6n+4}\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Prove by mathematical induction that }\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\ldots
\displaystyle +\frac{1}{(3n-2)(3n+1)}=\frac{n}{3n+1}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\ldots+\frac{1}{(3n-2)(3n+1)}=\frac{n}{3n+1}
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\frac{1}{1\times4}=\frac{1}{4}
\displaystyle \text{RHS}=\frac{1}{3(1)+1}=\frac{1}{4}
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\ldots+\frac{1}{(3k-2)(3k+1)}
\displaystyle =\frac{k}{3k+1}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=\frac{1}{1\times4}+\frac{1}{4\times7}+\ldots+\frac{1}{(3k-2)(3k+1)}
\displaystyle +\frac{1}{[3(k+1)-2][3(k+1)+1]}
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)}
\displaystyle =\frac{1}{3k+1}\left[k+\frac{1}{3k+4}\right]
\displaystyle =\frac{1}{3k+1}\left[\frac{k(3k+4)+1}{3k+4}\right]
\displaystyle =\frac{3k^2+4k+1}{(3k+1)(3k+4)}
\displaystyle =\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}
\displaystyle =\frac{k+1}{3k+4}
\displaystyle =\frac{k+1}{3(k+1)+1}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\ldots+\frac{1}{(3n-2)(3n+1)}
\displaystyle =\frac{n}{3n+1}\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Prove by mathematical induction that }\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\ldots
\displaystyle +\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\ldots+\frac{1}{(2n+1)(2n+3)}
\displaystyle =\frac{n}{3(2n+3)}
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\frac{1}{3\times5}=\frac{1}{15}
\displaystyle \text{RHS}=\frac{1}{3(2\times1+3)}=\frac{1}{15}
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\ldots+\frac{1}{(2k+1)(2k+3)}
\displaystyle =\frac{k}{3(2k+3)}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=\frac{1}{3\times5}+\frac{1}{5\times7}+\ldots+\frac{1}{(2k+1)(2k+3)}
\displaystyle +\frac{1}{(2k+3)(2k+5)}
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)}
\displaystyle =\frac{1}{2k+3}\left[\frac{k}{3}+\frac{1}{2k+5}\right]
\displaystyle =\frac{1}{2k+3}\left[\frac{k(2k+5)+3}{3(2k+5)}\right]
\displaystyle =\frac{2k^2+5k+3}{3(2k+3)(2k+5)}
\displaystyle =\frac{(2k+3)(k+1)}{3(2k+3)(2k+5)}
\displaystyle =\frac{k+1}{3(2k+5)}
\displaystyle =\frac{k+1}{3\{2(k+1)+3\}}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\ldots+\frac{1}{(2n+1)(2n+3)}
\displaystyle =\frac{n}{3(2n+3)}\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Prove by mathematical induction that }\frac{1}{3\times7}+\frac{1}{7\times11}+\frac{1}{11\times15}+\ldots
\displaystyle +\frac{1}{(4n-1)(4n+3)}=\frac{n}{3(4n+3)}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):\frac{1}{3\times7}+\frac{1}{7\times11}+\frac{1}{11\times15}+\ldots+\frac{1}{(4n-1)(4n+3)}
\displaystyle =\frac{n}{3(4n+3)}
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=\frac{1}{3\times7}=\frac{1}{21}
\displaystyle \text{RHS}=\frac{1}{3(4\times1+3)}=\frac{1}{21}
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \frac{1}{3\times7}+\frac{1}{7\times11}+\frac{1}{11\times15}+\ldots+\frac{1}{(4k-1)(4k+3)}
\displaystyle =\frac{k}{3(4k+3)}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=\frac{1}{3\times7}+\frac{1}{7\times11}+\ldots+\frac{1}{(4k-1)(4k+3)}
\displaystyle +\frac{1}{[4(k+1)-1][4(k+1)+3]}
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{k}{3(4k+3)}+\frac{1}{(4k+3)(4k+7)}
\displaystyle =\frac{1}{4k+3}\left[\frac{k}{3}+\frac{1}{4k+7}\right]
\displaystyle =\frac{1}{4k+3}\left[\frac{k(4k+7)+3}{3(4k+7)}\right]
\displaystyle =\frac{4k^2+7k+3}{3(4k+3)(4k+7)}
\displaystyle =\frac{(4k+3)(k+1)}{3(4k+3)(4k+7)}
\displaystyle =\frac{k+1}{3(4k+7)}
\displaystyle =\frac{k+1}{3[4(k+1)+3]}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle \frac{1}{3\times7}+\frac{1}{7\times11}+\frac{1}{11\times15}+\ldots+\frac{1}{(4n-1)(4n+3)}
\displaystyle =\frac{n}{3(4n+3)}\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Prove by mathematical induction that }1\times2+2\times2^2+3\times2^3+\ldots
\displaystyle +n\times2^n=(n-1)2^{n+1}+2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1\times2+2\times2^2+3\times2^3+\ldots+n\times2^n=(n-1)2^{n+1}+2
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1\times2=2
\displaystyle \text{RHS}=(1-1)2^{1+1}+2=2
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1\times2+2\times2^2+3\times2^3+\ldots+k\times2^k
\displaystyle =(k-1)2^{k+1}+2\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1\times2+2\times2^2+\ldots+k\times2^k+(k+1)2^{k+1}
\displaystyle \text{Using equation (i),}
\displaystyle =(k-1)2^{k+1}+2+(k+1)2^{k+1}
\displaystyle =\left[(k-1)+(k+1)\right]2^{k+1}+2
\displaystyle =2k\times2^{k+1}+2
\displaystyle =k2^{k+2}+2
\displaystyle =[(k+1)-1]2^{(k+1)+1}+2
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1\times2+2\times2^2+3\times2^3+\ldots+n\times2^n=(n-1)2^{n+1}+2
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Prove by mathematical induction that }2+5+8+11+\ldots+(3n-1)
\displaystyle =\frac{1}{2}n(3n+1).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):2+5+8+11+\ldots+(3n-1)=\frac{1}{2}n(3n+1)
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=2
\displaystyle \text{RHS}=\frac{1}{2}\times1(3\times1+1)=2
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 2+5+8+11+\ldots+(3k-1)=\frac{1}{2}k(3k+1)\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=2+5+8+11+\ldots+(3k-1)+[3(k+1)-1]
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{1}{2}k(3k+1)+(3k+2)
\displaystyle =\frac{3k^2+k+6k+4}{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 =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 2+5+8+11+\ldots+(3n-1)=\frac{1}{2}n(3n+1)\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Prove by mathematical induction that }1\times3+2\times4+3\times5+\ldots
\displaystyle +n(n+2)=\frac{1}{6}n(n+1)(2n+7).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1\times3+2\times4+3\times5+\ldots+n(n+2)=\frac{1}{6}n(n+1)(2n+7)
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1\times3=3
\displaystyle \text{RHS}=\frac{1}{6}\times1(1+1)(2\times1+7)=3
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1\times3+2\times4+3\times5+\ldots+k(k+2)
\displaystyle =\frac{1}{6}k(k+1)(2k+7)\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1\times3+2\times4+\ldots+k(k+2)+(k+1)(k+3)
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{1}{6}k(k+1)(2k+7)+(k+1)(k+3)
\displaystyle =(k+1)\left[\frac{k(2k+7)}{6}+(k+3)\right]
\displaystyle =\frac{k+1}{6}\left[k(2k+7)+6(k+3)\right]
\displaystyle =\frac{k+1}{6}\left(2k^2+13k+18\right)
\displaystyle =\frac{1}{6}(k+1)(2k+9)(k+2)
\displaystyle =\frac{1}{6}(k+1)[(k+1)+1][2(k+1)+7]
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1\times3+2\times4+3\times5+\ldots+n(n+2)=\frac{1}{6}n(n+1)(2n+7)
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Prove by mathematical induction that }1\times3+3\times5+5\times7+\ldots
\displaystyle +(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1\times3+3\times5+5\times7+\ldots+(2n-1)(2n+1)
\displaystyle =\frac{n(4n^2+6n-1)}{3}
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1\times3=3
\displaystyle \text{RHS}=\frac{1(4\times1^2+6\times1-1)}{3}=3
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1\times3+3\times5+5\times7+\ldots+(2k-1)(2k+1)
\displaystyle =\frac{k(4k^2+6k-1)}{3}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1\times3+3\times5+\ldots+(2k-1)(2k+1)
\displaystyle +[2(k+1)-1][2(k+1)+1]
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{k(4k^2+6k-1)}{3}+(2k+1)(2k+3)
\displaystyle =\frac{4k^3+6k^2-k+3(4k^2+8k+3)}{3}
\displaystyle =\frac{4k^3+18k^2+23k+9}{3}
\displaystyle =\frac{(k+1)(4k^2+14k+9)}{3}
\displaystyle =\frac{(k+1)[4(k+1)^2+6(k+1)-1]}{3}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1\times3+3\times5+5\times7+\ldots+(2n-1)(2n+1)
\displaystyle =\frac{n(4n^2+6n-1)}{3}\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Prove by mathematical induction that }1\times2+2\times3+3\times4+\ldots
\displaystyle +n(n+1)=\frac{n(n+1)(n+2)}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1\times2+2\times3+3\times4+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1\times2=2
\displaystyle \text{RHS}=\frac{1(1+1)(1+2)}{3}=2
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1\times2+2\times3+3\times4+\ldots+k(k+1)
\displaystyle =\frac{k(k+1)(k+2)}{3}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1\times2+2\times3+\ldots+k(k+1)+(k+1)(k+2)
\displaystyle \text{Using equation (i),}
\displaystyle =\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)
\displaystyle =(k+1)(k+2)\left(\frac{k}{3}+1\right)
\displaystyle =\frac{(k+1)(k+2)(k+3)}{3}
\displaystyle =\frac{(k+1)[(k+1)+1][(k+1)+2]}{3}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1\times2+2\times3+3\times4+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Prove by mathematical induction that }\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^n}
\displaystyle =1-\frac{1}{2^n}.
\displaystyle \text{Answer:}
\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,
\displaystyle \text{LHS}=\frac{1}{2}
\displaystyle \text{RHS}=1-\frac{1}{2}=\frac{1}{2}
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^k}
\displaystyle =1-\frac{1}{2^k}\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^k}+\frac{1}{2^{k+1}}
\displaystyle \text{Using equation (i),}
\displaystyle =1-\frac{1}{2^k}+\frac{1}{2^{k+1}}
\displaystyle =1-\frac{2}{2^{k+1}}+\frac{1}{2^{k+1}}
\displaystyle =1-\frac{1}{2^{k+1}}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{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}
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Prove by mathematical induction that }1^2+3^2+5^2+\ldots+(2n-1)^2
\displaystyle =\frac{1}{3}n(4n^2-1).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):1^2+3^2+5^2+\ldots+(2n-1)^2=\frac{1}{3}n(4n^2-1)
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=1^2=1
\displaystyle \text{RHS}=\frac{1}{3}\times1(4\times1^2-1)=1
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 1^2+3^2+5^2+\ldots+(2k-1)^2
\displaystyle =\frac{1}{3}k(4k^2-1)\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=1^2+3^2+5^2+\ldots+(2k-1)^2+[2(k+1)-1]^2
\displaystyle \text{Using equation (i),}
\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)\left[\frac{k(2k-1)}{3}+(2k+1)\right]
\displaystyle =\frac{2k+1}{3}\left[2k^2-k+6k+3\right]
\displaystyle =\frac{2k+1}{3}\left(2k^2+5k+3\right)
\displaystyle =\frac{1}{3}(2k+1)(2k+3)(k+1)
\displaystyle =\frac{k+1}{3}\left(4k^2+8k+3\right)
\displaystyle =\frac{k+1}{3}\left[4(k+1)^2-1\right]
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 1^2+3^2+5^2+\ldots+(2n-1)^2=\frac{1}{3}n(4n^2-1)
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Prove by mathematical induction that }a+ar+ar^2+\ldots+ar^{n-1}
\displaystyle =a\left(\frac{r^n-1}{r-1}\right),\text{ where }r\neq1.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):a+ar+ar^2+\ldots+ar^{n-1}=a\left(\frac{r^n-1}{r-1}\right),\ r\neq1
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=a
\displaystyle \text{RHS}=a\left(\frac{r^1-1}{r-1}\right)=a
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore a+ar+ar^2+\ldots+ar^{k-1}=a\left(\frac{r^k-1}{r-1}\right)\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=a+ar+ar^2+\ldots+ar^{k-1}+ar^k
\displaystyle \text{Using equation (i),}
\displaystyle =a\left(\frac{r^k-1}{r-1}\right)+ar^k
\displaystyle =\frac{a(r^k-1)+ar^k(r-1)}{r-1}
\displaystyle =\frac{ar^k-a+ar^{k+1}-ar^k}{r-1}
\displaystyle =\frac{ar^{k+1}-a}{r-1}
\displaystyle =a\left(\frac{r^{k+1}-1}{r-1}\right)
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle a+ar+ar^2+\ldots+ar^{n-1}=a\left(\frac{r^n-1}{r-1}\right),\ r\neq1,
\displaystyle \text{for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Prove by mathematical induction that }a+(a+d)+(a+2d)+\ldots
\displaystyle +(a+(n-1)d)=\frac{n}{2}[2a+(n-1)d].
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):a+(a+d)+(a+2d)+\ldots+(a+(n-1)d)
\displaystyle =\frac{n}{2}[2a+(n-1)d]
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=a
\displaystyle \text{RHS}=\frac{1}{2}[2a+(1-1)d]=a
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore a+(a+d)+(a+2d)+\ldots+(a+(k-1)d)
\displaystyle =\frac{k}{2}[2a+(k-1)d]\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle \text{LHS of }P(k+1)=a+(a+d)+\ldots+(a+(k-1)d)+(a+kd)
\displaystyle \text{Using equation (i),}
\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 =a(k+1)+\frac{k(k+1)d}{2}
\displaystyle =\frac{2a(k+1)+k(k+1)d}{2}
\displaystyle =\frac{k+1}{2}(2a+kd)
\displaystyle =\frac{k+1}{2}[2a+\{(k+1)-1\}d]
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle a+(a+d)+(a+2d)+\ldots+(a+(n-1)d)
\displaystyle =\frac{n}{2}[2a+(n-1)d]\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Prove by mathematical induction that }5^{2n}-1\text{ is divisible by }24
\displaystyle \text{for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):5^{2n}-1\text{ is divisible by }24.
\displaystyle \text{For }n=1,
\displaystyle 5^{2\times1}-1=25-1=24
\displaystyle \text{Since }24\text{ is divisible by }24,\ P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 5^{2k}-1\text{ is divisible by }24.
\displaystyle \therefore 5^{2k}-1=24\lambda,\text{ where }\lambda\in N
\displaystyle \therefore 5^{2k}=24\lambda+1\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle 5^{2(k+1)}-1=5^{2k+2}-1
\displaystyle =25\times5^{2k}-1
\displaystyle \text{Using equation (i),}
\displaystyle =25(24\lambda+1)-1
\displaystyle =25\times24\lambda+24
\displaystyle =24(25\lambda+1)
\displaystyle \text{Let }\mu=25\lambda+1.
\displaystyle \therefore 5^{2(k+1)}-1=24\mu
\displaystyle \therefore 5^{2(k+1)}-1\text{ is divisible by }24.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 5^{2n}-1\text{ is divisible by }24\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Prove by mathematical induction that }3^{2n}+7\text{ is divisible by }8
\displaystyle \text{for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):3^{2n}+7\text{ is divisible by }8.
\displaystyle \text{For }n=1,
\displaystyle 3^{2\times1}+7=3^2+7=9+7=16
\displaystyle \text{Since }16\text{ is divisible by }8,\ P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 3^{2k}+7\text{ is divisible by }8.
\displaystyle \therefore 3^{2k}+7=8\lambda,\text{ where }\lambda\in Z
\displaystyle \therefore 3^{2k}=8\lambda-7\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle 3^{2(k+1)}+7=3^{2k+2}+7
\displaystyle =9\times3^{2k}+7
\displaystyle \text{Using equation (i),}
\displaystyle =9(8\lambda-7)+7
\displaystyle =72\lambda-63+7
\displaystyle =72\lambda-56
\displaystyle =8(9\lambda-7)
\displaystyle \text{Let }\mu=9\lambda-7.
\displaystyle \therefore 3^{2(k+1)}+7=8\mu
\displaystyle \therefore 3^{2(k+1)}+7\text{ is divisible by }8.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 3^{2n}+7\text{ is divisible by }8\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Prove by mathematical induction that }5^{2n+2}-24n-25\text{ is divisible}
\displaystyle \text{by }576\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):5^{2n+2}-24n-25\text{ is divisible by }576.
\displaystyle \text{For }n=1,
\displaystyle 5^{2(1)+2}-24(1)-25=5^4-24-25
\displaystyle =625-49=576
\displaystyle \text{Since }576\text{ is divisible by }576,\ P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 5^{2k+2}-24k-25\text{ is divisible by }576.
\displaystyle \therefore 5^{2k+2}-24k-25=576\lambda,\text{ where }\lambda\in Z
\displaystyle \therefore 5^{2k+2}=576\lambda+24k+25\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle 5^{2(k+1)+2}-24(k+1)-25
\displaystyle =5^{2k+4}-24k-24-25
\displaystyle =25\times5^{2k+2}-24k-49
\displaystyle \text{Using equation (i),}
\displaystyle =25(576\lambda+24k+25)-24k-49
\displaystyle =25\times576\lambda+600k+625-24k-49
\displaystyle =25\times576\lambda+576k+576
\displaystyle =576(25\lambda+k+1)
\displaystyle \text{Let }\mu=25\lambda+k+1.
\displaystyle \therefore 5^{2(k+1)+2}-24(k+1)-25=576\mu
\displaystyle \therefore 5^{2(k+1)+2}-24(k+1)-25\text{ is divisible by }576.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 5^{2n+2}-24n-25\text{ is divisible by }576\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Prove by mathematical induction that }3^{2n+2}-8n-9\text{ is divisible by }8
\displaystyle \text{for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):3^{2n+2}-8n-9\text{ is divisible by }8.
\displaystyle \text{For }n=1,
\displaystyle 3^{2(1)+2}-8(1)-9=3^4-8-9
\displaystyle =81-17=64
\displaystyle \text{Since }64\text{ is divisible by }8,\ P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 3^{2k+2}-8k-9\text{ is divisible by }8.
\displaystyle \therefore 3^{2k+2}-8k-9=8\lambda,\text{ where }\lambda\in Z
\displaystyle \therefore 3^{2k+2}=8\lambda+8k+9\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle 3^{2(k+1)+2}-8(k+1)-9
\displaystyle =3^{2k+4}-8k-8-9
\displaystyle =9\times3^{2k+2}-8k-17
\displaystyle \text{Using equation (i),}
\displaystyle =9(8\lambda+8k+9)-8k-17
\displaystyle =72\lambda+72k+81-8k-17
\displaystyle =72\lambda+64k+64
\displaystyle =8(9\lambda+8k+8)
\displaystyle \text{Let }\mu=9\lambda+8k+8.
\displaystyle \therefore 3^{2(k+1)+2}-8(k+1)-9=8\mu
\displaystyle \therefore 3^{2(k+1)+2}-8(k+1)-9\text{ is divisible by }8.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 3^{2n+2}-8n-9\text{ is divisible by }8\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Prove by mathematical induction that }(ab)^n=a^nb^n\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):(ab)^n=a^nb^n
\displaystyle \text{For }n=1,
\displaystyle \text{LHS}=ab
\displaystyle \text{RHS}=a^1b^1=ab
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore (ab)^k=a^kb^k\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle (ab)^{k+1}=(ab)^k(ab)
\displaystyle \text{Using equation (i),}
\displaystyle =(a^kb^k)(ab)
\displaystyle =a^ka\,b^kb
\displaystyle =a^{k+1}b^{k+1}
\displaystyle =\text{RHS of }P(k+1)
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle (ab)^n=a^nb^n\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Prove by mathematical induction that }n(n+1)(n+5)\text{ is a multiple of }3
\displaystyle \text{for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):n(n+1)(n+5)\text{ is a multiple of }3.
\displaystyle \text{For }n=1,
\displaystyle 1(1+1)(1+5)=1\times2\times6=12
\displaystyle \text{Since }12\text{ is divisible by }3,\ P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore k(k+1)(k+5)\text{ is divisible by }3.
\displaystyle \therefore k(k+1)(k+5)=3\lambda,\text{ where }\lambda\in Z
\displaystyle \therefore k^3+6k^2+5k=3\lambda\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle (k+1)[(k+1)+1][(k+1)+5]
\displaystyle =(k+1)(k+2)(k+6)
\displaystyle =(k^2+3k+2)(k+6)
\displaystyle =k^3+9k^2+20k+12
\displaystyle =(k^3+6k^2+5k)+(3k^2+15k+12)
\displaystyle \text{Using equation (i),}
\displaystyle =3\lambda+3(k^2+5k+4)
\displaystyle =3[\lambda+k^2+5k+4]
\displaystyle \text{Let }\mu=\lambda+k^2+5k+4.
\displaystyle \therefore (k+1)(k+2)(k+6)=3\mu
\displaystyle \therefore (k+1)(k+2)(k+6)\text{ is divisible by }3.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle n(n+1)(n+5)\text{ is a multiple of }3\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Prove by mathematical induction that }7^{2n}+2^{3n-3}\times3^{n-1}
\displaystyle \text{is divisible by }25\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n):7^{2n}+2^{3n-3}\times3^{n-1}\text{ is divisible by }25.
\displaystyle \text{For }n=1,
\displaystyle 7^{2(1)}+2^{3(1)-3}\times3^{1-1}
\displaystyle =7^2+2^0\times3^0
\displaystyle =49+1=50
\displaystyle \text{Since }50\text{ is divisible by }25,\ P(1)\text{ is true.}
\displaystyle \text{Assume that }P(k)\text{ is true for some }k\in N.
\displaystyle \therefore 7^{2k}+2^{3k-3}\times3^{k-1}\text{ is divisible by }25.
\displaystyle \therefore 7^{2k}+2^{3k-3}\times3^{k-1}=25\lambda,\text{ where }\lambda\in Z
\displaystyle \qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{We shall now prove that }P(k+1)\text{ is true.}
\displaystyle 7^{2(k+1)}+2^{3(k+1)-3}\times3^{(k+1)-1}
\displaystyle =49\times7^{2k}+2^{3k}\times3^k
\displaystyle =49\times7^{2k}+24\left(2^{3k-3}\times3^{k-1}\right)
\displaystyle =49\left[7^{2k}+2^{3k-3}\times3^{k-1}\right]
\displaystyle \quad-25\left(2^{3k-3}\times3^{k-1}\right)
\displaystyle \text{Using equation (i),}
\displaystyle =49(25\lambda)-25\left(2^{3k-3}\times3^{k-1}\right)
\displaystyle =25\left[49\lambda-2^{3k-3}\times3^{k-1}\right]
\displaystyle \text{Let }\mu=49\lambda-2^{3k-3}\times3^{k-1}.
\displaystyle \therefore 7^{2(k+1)}+2^{3(k+1)-3}\times3^{(k+1)-1}=25\mu
\displaystyle \therefore 7^{2(k+1)}+2^{3(k+1)-3}\times3^{(k+1)-1}\text{ is divisible by }25.
\displaystyle \therefore P(k+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle 7^{2n}+2^{3n-3}\times3^{n-1}\text{ is divisible by }25\text{ for all }n\in N.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.