Question

1.2 + 2.3 + 3.4 + ... + n (n + 1) = $\frac{n(n+1)(n+2)}{3}$

Answer 1

Let P(n) be the given statement.
Now,
$$\begin{gathered} P\left(n\right)=1.2+2.3+3.4+...+n(n+1)=\frac{n(n+1)(n+2)}{3} \\ \mathrm{Step}1: \\ P\left(1\right)=1.2=2=\frac{1(1+1)(1+2)}{3} \\ \mathrm{Hence},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)\mathrm{be}\mathrm{true}. \\ \mathrm{Then}, \\ 1.2+2.3+...+m(m+1)=\frac{m(m+1)(m+2)}{3} \\ \mathrm{To}\mathrm{prove}:P(m+1)\mathrm{is}\mathrm{true}. \\ \mathrm{That}\mathrm{is}, \\ 1.2+2.3+...+(m+1)(m+2)=\frac{(m+1)(m+2)(m+3)}{3} \\ \mathrm{Now},P\left(m\right)\mathrm{is} \\ 1.2+2.3+...+m(m+1)=\frac{m(m+1)(m+2)}{3} \\ \Rightarrow 1.2+2.3+...+m(m+1)+(m+1)(m+2)=\frac{m(m+1)(m+2)}{3}+(m+1)(m+2) \\ \Rightarrow P(m+1)=\frac{(m+1)(m+2)(m+3)}{3} \\ \mathrm{Thus},P(m+1)\mathrm{is}\mathrm{true}. \\ By\mathrm{the}p\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}\mathrm{induction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N.\text{'} \end{gathered}$$