Question

1.3 + 2.4 + 3.5 + ... + n. (n + 2) =$\frac{1}{6}n(n+1)(2n+7)$

Answer 1

Let P(n) be the given statement.
Now,
$$\begin{gathered} P\left(n\right)=1.3+2.4+3.5+...+n.(n+2)=\frac{1}{6}n(n+1)(2n+7) \\ \mathrm{Step}1: \\ P\left(1\right)=1.3=3=\frac{1}{6}\times 1(1+1)(2\times 1+7) \\ \mathrm{Hence},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)\mathrm{be}\mathrm{true}. \\ Then, \\ 1.3+2.4+...+m.(m+2)=\frac{1}{6}m(m+1)(2m+7) \\ To\mathrm{prove}:P(m+1)\mathrm{is}\mathrm{true}. \\ \mathrm{That}\mathrm{is}, \\ 1.3+2.4+...+(m+1)(m+3)=\frac{1}{6}(m+1)(m+2)(2m+9) \\ P\left(m\right)\mathrm{is}\mathrm{equal}\mathrm{to}1.3+2.4+...+m(m+2)=\frac{1}{6}m(m+1)(2m+7). \\ Thus,wehave: \\ 1.3+2.4+...+m(m+2)+(m+1)(m+3)=\frac{1}{6}m(m+1)(2m+7)+(m+1)(m+3)\left[\mathrm{Adding}(m+1)(m+3)\mathrm{to}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow 1.3+2.4+...+(m+1)(m+3)=\frac{1}{6}(m+1)\left[2m^2+7m+6m+18\right] \\ =\frac{1}{6}(m+1)(2m^2+13m+18) \\ =\frac{1}{6}(m+1)(2m+9)(m+2) \\ \mathrm{Thus},P(m+1)istrue. \\ Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}\mathrm{induction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N. \end{gathered}$$