\displaystyle \textbf{Question 1: }\text{Let }P(n)\text{ be the statement that }7\text{ divides }(2^{3n}-1).
\displaystyle \text{What is }P(n+1)?
\displaystyle \text{Answer:}
\displaystyle P(n+1)\text{ is the statement that }7\text{ divides }\left(2^{3(n+1)}-1\right).
\displaystyle \text{This is obtained by replacing }n\text{ with }(n+1)\text{ in the statement }P(n).
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }P(n)\text{ is the statement }n^2>100,\text{ prove that whenever}
\displaystyle P(r)\text{ is true, }P(r+1)\text{ is also true.}
\displaystyle \text{Answer:}
\displaystyle P(n)\text{ is the statement }n^2>100.
\displaystyle \text{Assume that }P(r)\text{ is true.}
\displaystyle \therefore r^2>100.
\displaystyle \text{We have to prove that }P(r+1)\text{ is true, i.e.}
\displaystyle (r+1)^2>100.
\displaystyle \text{Now,}
\displaystyle r^2>100
\displaystyle \Rightarrow r^2+2r+1>100+2r+1.
\displaystyle \Rightarrow (r+1)^2>100+2r+1.
\displaystyle \text{Since }r\in N,\ 2r+1>0.
\displaystyle \therefore 100+2r+1>100.
\displaystyle \therefore (r+1)^2>100.
\displaystyle \therefore P(r+1)\text{ is true.}
\displaystyle \text{Hence, whenever }P(r)\text{ is true, }P(r+1)\text{ is also true.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Let }P(n)\text{ be the statement }3^n>n.
\displaystyle \text{If }P(n)\text{ is true, prove that }P(n+1)\text{ is true.}
\displaystyle \text{Answer:}
\displaystyle \text{Assume that }P(n)\text{ is true.}
\displaystyle \therefore 3^n>n.
\displaystyle \text{We have to prove that }P(n+1)\text{ is true, i.e.}
\displaystyle 3^{\,n+1}>n+1.
\displaystyle \text{Now,}
\displaystyle 3^n>n
\displaystyle \Rightarrow 3\cdot3^n>3n.
\displaystyle \Rightarrow 3^{\,n+1}>3n.
\displaystyle \text{Since }n\in N,\ 2n\ge2>1.
\displaystyle \therefore 3n=n+2n>n+1.
\displaystyle \therefore 3^{\,n+1}>n+1.
\displaystyle \therefore P(n+1)\text{ is true.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }P(n)\text{ is the statement }2^{3n}-1\text{ is an integral multiple of }7,
\displaystyle \text{and if }P(r)\text{ is true, prove that }P(r+1)\text{ is true.}
\displaystyle \text{Answer:}
\displaystyle \text{Assume that }P(r)\text{ is true.}
\displaystyle \therefore 2^{3r}-1\text{ is an integral multiple of }7.
\displaystyle \therefore 2^{3r}-1=7\lambda,\text{ for some }\lambda\in\mathbb{Z}.
\displaystyle \therefore 2^{3r}=7\lambda+1.
\displaystyle \text{We have to prove that }P(r+1)\text{ is true.}
\displaystyle \text{Now,}
\displaystyle 2^{3(r+1)}-1=2^{3r}\cdot2^3-1
\displaystyle =(7\lambda+1)\cdot8-1
\displaystyle =56\lambda+8-1
\displaystyle =56\lambda+7
\displaystyle =7(8\lambda+1).
\displaystyle \text{Let }\mu=8\lambda+1.
\displaystyle \therefore 2^{3(r+1)}-1=7\mu,\text{ where }\mu\in\mathbb{Z}.
\displaystyle \therefore 2^{3(r+1)}-1\text{ is an integral multiple of }7.
\displaystyle \therefore P(r+1)\text{ is true.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Prove by the principle of mathematical induction that }n^2+n
\displaystyle \text{is an even natural number for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that }n^2+n\text{ is an even natural number.}
\displaystyle \text{For }n=1,
\displaystyle 1^2+1=2,
\displaystyle \text{which is an even natural number.}
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(m)\text{ is true for some }m\in N.
\displaystyle \therefore m^2+m\text{ is even.}
\displaystyle \therefore m^2+m=2\lambda,\text{ for some }\lambda\in N.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=m+1.
\displaystyle (m+1)^2+(m+1)
\displaystyle =m^2+2m+1+m+1
\displaystyle =(m^2+m)+2m+2
\displaystyle =(m^2+m)+2(m+1)
\displaystyle =2\lambda+2(m+1)\qquad\text{[Using (i)]}
\displaystyle =2(\lambda+m+1).
\displaystyle \text{Since }\lambda+m+1\in N,
\displaystyle (m+1)^2+(m+1)\text{ is an even natural number.}
\displaystyle \therefore P(m+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle n^2+n\text{ is an even natural number for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Prove by the principle of mathematical induction that}
\displaystyle n(n+1)(2n+1)\text{ is divisible by }6\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that }n(n+1)(2n+1)\text{ is divisible by }6.
\displaystyle \text{For }n=1,
\displaystyle 1(1+1)(2\cdot1+1)=1\cdot2\cdot3=6.
\displaystyle \text{Since }6\text{ is divisible by }6,\ P(1)\text{ is true.}
\displaystyle \text{Assume that }P(m)\text{ is true for some }m\in N.
\displaystyle \therefore m(m+1)(2m+1)\text{ is divisible by }6.
\displaystyle \therefore m(m+1)(2m+1)=6\lambda,\text{ for some }\lambda\in N.\qquad\text{...(i)}
\displaystyle \text{Now, consider the expression for }n=m+1.
\displaystyle (m+1)\{(m+1)+1\}\{2(m+1)+1\}
\displaystyle =(m+1)(m+2)(2m+3)
\displaystyle =(m+1)(m+2)\{(2m+1)+2\}
\displaystyle =(m+1)(m+2)(2m+1)+2(m+1)(m+2)
\displaystyle =m(m+1)(2m+1)+2(m+1)(2m+1)+2(m+1)(m+2)
\displaystyle =m(m+1)(2m+1)+2(m+1)\{2m+1+m+2\}
\displaystyle =m(m+1)(2m+1)+2(m+1)(3m+3)
\displaystyle =m(m+1)(2m+1)+6(m+1)^2
\displaystyle =6\lambda+6(m+1)^2\qquad\text{[Using (i)]}
\displaystyle =6\{\lambda+(m+1)^2\}.
\displaystyle \text{Since }\lambda+(m+1)^2\in N,
\displaystyle (m+1)(m+2)(2m+3)\text{ is divisible by }6.
\displaystyle \therefore P(m+1)\text{ is true.}
\displaystyle \text{Hence, by the principle of mathematical induction,}
\displaystyle n(n+1)(2n+1)\text{ is divisible by }6\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Prove by induction that the expression }S_n=n^3+3n^2+5n+3\text{ is}
\displaystyle \text{divisible by }3\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that }S_n=n^3+3n^2+5n+3\text{ is divisible by }3.
\displaystyle \text{For }n=1,
\displaystyle S_1=1^3+3(1)^2+5(1)+3=1+3+5+3=12
\displaystyle \text{Since }12\text{ is divisible by }3,\text{ therefore }P(1)\text{ is true.}
\displaystyle \text{Assume that }P(m)\text{ is true for some }m\in N.
\displaystyle \therefore S_m=m^3+3m^2+5m+3\text{ is divisible by }3.
\displaystyle \therefore m^3+3m^2+5m+3=3\lambda\text{ for some }\lambda\in N.
\displaystyle \text{We shall now prove that }P(m+1)\text{ is true.}
\displaystyle S_{m+1}=(m+1)^3+3(m+1)^2+5(m+1)+3
\displaystyle =m^3+3m^2+5m+3+3m^2+9m+9
\displaystyle =3\lambda+3(m^2+3m+3)
\displaystyle =3(\lambda+m^2+3m+3)
\displaystyle \text{Let }\mu=\lambda+m^2+3m+3.
\displaystyle \text{Since }\lambda,m\in N,\text{ we have }\mu\in N.
\displaystyle \therefore S_{m+1}=3\mu
\displaystyle \therefore S_{m+1}\text{ is divisible by }3.
\displaystyle \therefore P(m+1)\text{ is true.}
\displaystyle \text{Thus, }P(m)\text{ is true }\Rightarrow P(m+1)\text{ is true.}
\displaystyle \therefore \text{By the principle of mathematical induction, }S_n=n^3+3n^2+5n+3
\displaystyle \text{is divisible by }3\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Using the principle of mathematical induction, prove that}
\displaystyle x^{2n}-y^{2n}\text{ is divisible by }x+y\text{ for all }n\in N,\text{ where }x,y\in Z.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that }x^{2n}-y^{2n}\text{ is divisible by }x+y.
\displaystyle \text{For }n=1,
\displaystyle x^2-y^2=(x-y)(x+y)
\displaystyle \therefore x^2-y^2\text{ is divisible by }x+y.
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(m)\text{ is true for some }m\in N.
\displaystyle \therefore x^{2m}-y^{2m}\text{ is divisible by }x+y.
\displaystyle \therefore x^{2m}-y^{2m}=\lambda(x+y)\text{ for some }\lambda\in Z.
\displaystyle \text{We shall now prove that }P(m+1)\text{ is true.}
\displaystyle x^{2m+2}-y^{2m+2}
\displaystyle =x^{2m+2}-x^{2m}y^2+x^{2m}y^2-y^{2m+2}
\displaystyle =x^{2m}(x^2-y^2)+y^2(x^{2m}-y^{2m})
\displaystyle =x^{2m}(x-y)(x+y)+y^2\lambda(x+y)
\displaystyle =(x+y)\left[x^{2m}(x-y)+\lambda y^2\right]
\displaystyle \therefore x^{2m+2}-y^{2m+2}\text{ is divisible by }x+y.
\displaystyle \therefore P(m+1)\text{ is true.}
\displaystyle \text{Thus, }P(m)\text{ is true }\Rightarrow P(m+1)\text{ is true.}
\displaystyle \therefore \text{By the principle of mathematical induction, }x^{2n}-y^{2n}
\displaystyle \text{is divisible by }x+y\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Using the principle of mathematical induction, prove that}
\displaystyle (2^{3n}-1)\text{ is divisible by }7\text{ for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that }2^{3n}-1\text{ is divisible by }7.
\displaystyle \text{For }n=1,
\displaystyle 2^{3\times1}-1=8-1=7
\displaystyle \text{Since }7\text{ is divisible by }7,\text{ therefore }P(1)\text{ is true.}
\displaystyle \text{Assume that }P(m)\text{ is true for some }m\in N.
\displaystyle \therefore 2^{3m}-1\text{ is divisible by }7.
\displaystyle \therefore 2^{3m}-1=7\lambda\text{ for some }\lambda\in N.
\displaystyle \text{We shall now prove that }P(m+1)\text{ is true.}
\displaystyle 2^{3(m+1)}-1=2^{3m}\times2^3-1
\displaystyle =8(2^{3m})-1
\displaystyle =8(7\lambda+1)-1
\displaystyle =56\lambda+8-1
\displaystyle =7(8\lambda+1)
\displaystyle \therefore 2^{3(m+1)}-1\text{ is divisible by }7.
\displaystyle \therefore P(m+1)\text{ is true.}
\displaystyle \text{Thus, }P(m)\text{ is true }\Rightarrow P(m+1)\text{ is true.}
\displaystyle \therefore \text{By the principle of mathematical induction, }2^{3n}-1
\displaystyle \text{is divisible by }7\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Prove that }2\cdot7^n+3\cdot5^n-5\text{ is divisible by }24
\displaystyle \text{for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that }2\cdot7^n+3\cdot5^n-5\text{ is divisible by }24.
\displaystyle \text{For }n=1,
\displaystyle 2\cdot7^1+3\cdot5^1-5=14+15-5=24
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(m)\text{ is true for some }m\in N.
\displaystyle \therefore 2\cdot7^m+3\cdot5^m-5=24\lambda\text{ for some }\lambda\in N.
\displaystyle \therefore 3\cdot5^m=24\lambda+5-2\cdot7^m
\displaystyle \text{We shall now prove that }P(m+1)\text{ is true.}
\displaystyle 2\cdot7^{m+1}+3\cdot5^{m+1}-5
\displaystyle =2\cdot7^{m+1}+5(3\cdot5^m)-5
\displaystyle =2\cdot7^{m+1}+5(24\lambda+5-2\cdot7^m)-5
\displaystyle =2\cdot7^{m+1}+120\lambda+25-10\cdot7^m-5
\displaystyle =(14-10)7^m+120\lambda+20
\displaystyle =4\cdot7^m+120\lambda+20
\displaystyle =4(7^m-1)+24(5\lambda+1)
\displaystyle \text{Now, }7^m-1=(7-1)(7^{m-1}+7^{m-2}+\cdots+7+1)
\displaystyle \therefore 7^m-1\text{ is divisible by }6.
\displaystyle \therefore 7^m-1=6\mu\text{ for some }\mu\in N.
\displaystyle \therefore 2\cdot7^{m+1}+3\cdot5^{m+1}-5
\displaystyle =4(6\mu)+24(5\lambda+1)
\displaystyle =24(\mu+5\lambda+1)
\displaystyle \therefore 2\cdot7^{m+1}+3\cdot5^{m+1}-5\text{ is divisible by }24.
\displaystyle \therefore P(m+1)\text{ is true.}
\displaystyle \text{Thus, }P(m)\text{ is true }\Rightarrow P(m+1)\text{ is true.}
\displaystyle \therefore \text{By the principle of mathematical induction, }2\cdot7^n+3\cdot5^n-5
\displaystyle \text{is divisible by }24\text{ for all }n\in N.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Prove by induction that }2n+7<(n+3)^2\text{ for all natural}
\displaystyle \text{numbers }n.\text{ Using this, prove by induction that }(n+3)^2\leq2^{n+3}
\displaystyle \text{for all }n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(n)\text{ be the statement that }2n+7<(n+3)^2.
\displaystyle \text{For }n=1,
\displaystyle 2(1)+7<(1+3)^2
\displaystyle \Rightarrow 9<16
\displaystyle \therefore P(1)\text{ is true.}
\displaystyle \text{Assume that }P(m)\text{ is true for some }m\in N.
\displaystyle \therefore 2m+7<(m+3)^2
\displaystyle \text{Adding }2\text{ to both sides,}
\displaystyle 2m+9<(m+3)^2+2
\displaystyle \Rightarrow 2(m+1)+7<m^2+6m+11
\displaystyle \text{Since }m^2+6m+11<m^2+8m+16\text{ for }m\in N,
\displaystyle 2(m+1)+7<m^2+8m+16
\displaystyle \Rightarrow 2(m+1)+7<(m+4)^2
\displaystyle \Rightarrow 2(m+1)+7<[(m+1)+3]^2
\displaystyle \therefore P(m+1)\text{ is true.}
\displaystyle \text{Thus, }P(m)\text{ is true }\Rightarrow P(m+1)\text{ is true.}
\displaystyle \therefore \text{By the principle of mathematical induction, }2n+7<(n+3)^2
\displaystyle \text{for all }n\in N.
\displaystyle \text{Now, let }P'(n)\text{ be the statement that }(n+3)^2\leq2^{n+3}.
\displaystyle \text{For }n=1,
\displaystyle (1+3)^2\leq2^{1+3}
\displaystyle \Rightarrow 16\leq16
\displaystyle \therefore P'(1)\text{ is true.}
\displaystyle \text{Assume that }P'(m)\text{ is true for some }m\in N.
\displaystyle \therefore (m+3)^2\leq2^{m+3}
\displaystyle \text{We shall now prove that }P'(m+1)\text{ is true.}
\displaystyle (m+4)^2=(m+3)^2+2m+7
\displaystyle \text{From the result proved above, }2m+7<(m+3)^2.
\displaystyle \therefore (m+4)^2<(m+3)^2+(m+3)^2
\displaystyle =2(m+3)^2
\displaystyle \leq2\cdot2^{m+3}
\displaystyle =2^{m+4}
\displaystyle \therefore (m+4)^2\leq2^{m+4}
\displaystyle \Rightarrow [(m+1)+3]^2\leq2^{(m+1)+3}
\displaystyle \therefore P'(m+1)\text{ is true.}
\displaystyle \text{Thus, }P'(m)\text{ is true }\Rightarrow P'(m+1)\text{ is true.}
\displaystyle \therefore \text{By the principle of mathematical induction, }(n+3)^2\leq2^{n+3}
\displaystyle \text{for all }n\in N.
\displaystyle \\

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