Using mathematical induction prove 2.3+3.4+4.5+..upto n terms = n (n^2+6n+11)/3.

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$$\begin{gathered} \mathrm{We} \mathrm{have} 2.3+3.4+4.5+... \\ \mathrm{Clearly} {\mathrm{n}}_{\mathrm{th}} \mathrm{term} \mathrm{will} \mathrm{be} \left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right) \\ \mathrm{P}\left(\mathrm{n}\right): 2.3+3.4+4.5+...+\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)=\frac{\mathrm{n}\left({\mathrm{n}}^2+6\mathrm{n}+11\right)}{3} \\ \mathbf{Step}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathbf{Base}\mathbf{ }\mathbf{Case} \\ \mathrm{For} \mathrm{n}=1 \\ 2.3=\frac{1\left(1+6+11\right)}{3}=\frac{18}{3} \\ \Rightarrow 6=6 \\ \mathrm{Hence} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=1 \\ \mathbf{Step}\mathbf{ }\mathbf{2}\mathbf{:}\mathbf{ }\mathbf{Inductive}\mathbf{ }\mathbf{Hypothesis} \\ \mathrm{Let} \mathrm{P}\left(\mathrm{n}\right) \mathrm{be} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k} \\ \Rightarrow 2.3+3.4+4.5+...+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)=\frac{\mathrm{k}\left({\mathrm{k}}^2+6\mathrm{k}+11\right)}{3} ;\left(\mathrm{i}\right) \\ \mathbf{Step}\mathbf{ }\mathbf{3}\mathbf{:}\mathbf{ }\mathbf{Inductive}\mathbf{ }\mathbf{Case} \\ \mathrm{Consider} \\ 2.3+3.4+4.5+...+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)+\left(\mathrm{k}+1+1\right)\left(\mathrm{k}+1+2\right) \\ =\frac{\mathrm{k}\left({\mathrm{k}}^2+6\mathrm{k}+11\right)}{3}+\left(\mathrm{k}+2\right)\left(\mathrm{k}+3\right) \left[\mathrm{Using} \left(\mathrm{i}\right)\right] \\ =\frac{\mathrm{k}\left({\mathrm{k}}^2+6\mathrm{k}+11\right)}{3}+\left({\mathrm{k}}^2+5\mathrm{k}+6\right) \\ =\frac{{\mathrm{k}}^3+6{\mathrm{k}}^2+11\mathrm{k}+3{\mathrm{k}}^2+15\mathrm{k}+18}{3} \\ =\frac{{\mathrm{k}}^3+6{\mathrm{k}}^2+11\mathrm{k}+3{\mathrm{k}}^2+15\mathrm{k}+1+11+6}{3} \\ =\frac{\left({\mathrm{k}}^3+1+3{\mathrm{k}}^2+3\mathrm{k}\right)+\left(6{\mathrm{k}}^2+12\mathrm{k}+6\right)+\left(11\mathrm{k}+11\right)}{3} \\ =\frac{\left({\mathrm{k}}^3+1^3+3{\mathrm{k}}^2+3\mathrm{k}\right)+6\left({\mathrm{k}}^2+2\mathrm{k}+1\right)+11\left(\mathrm{k}+1\right)}{3} \\ =\frac{{\left(\mathrm{k}+1\right)}^3+6{\left(\mathrm{k}+1\right)}^2+11\left(\mathrm{k}+1\right)}{3} \\ =\left(\mathrm{k}+1\right)\left\{\frac{{\left(\mathrm{k}+1\right)}^2+6\left(\mathrm{k}+1\right)+11}{3}\right\} \\ \Rightarrow 2.3+3.4+4.5+...+\left(\mathrm{k}+1\right)\left(\mathrm{k}+2\right)+\left(\mathrm{k}+1+1\right)\left(\mathrm{k}+1+2\right)=\left(\mathrm{k}+1\right)\left\{\frac{{\left(\mathrm{k}+1\right)}^2+6\left(\mathrm{k}+1\right)+11}{3}\right\} \\ \mathrm{Hence} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k}+1 \\ \mathbf{Since}\mathbf{}\mathbf{P}\left(\mathbf{n}\right)\mathbf{}\mathbf{is}\mathbf{}\mathbf{true}\mathbf{}\mathbf{for}\mathbf{}\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{and}\mathbf{}\mathbf{assumption}\mathbf{}\mathbf{of}\mathbf{}\mathbf{truth}\mathbf{}\mathbf{for}\mathbf{}\mathbf{n}\mathbf{=}\mathbf{k}\mathbf{,}\mathbf{}\mathbf{gives}\mathbf{}\mathbf{us}\mathbf{}\mathbf{the}\mathbf{}\mathbf{result} \\ \mathbf{that}\mathbf{}\mathbf{P}\left(\mathbf{n}\right)\mathbf{}\mathbf{is}\mathbf{}\mathbf{also}\mathbf{}\mathbf{true}\mathbf{}\mathbf{for}\mathbf{}\mathbf{n}\mathbf{=}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{.} \\ \mathbf{Hence}\mathbf{}\mathbf{by}\mathbf{}\mathbf{Principal}\mathbf{}\mathbf{of}\mathbf{}\mathbf{mathematical}\mathbf{}\mathbf{induction}\mathbf{}\mathbf{P}\left(\mathbf{n}\right)\mathbf{}\mathbf{is}\mathbf{}\mathbf{true}\mathbf{ }\mathbf{for}\mathbf{ }\mathbf{all}\mathbf{ }\mathbf{n}\mathbf{\in }\mathbf{N} \end{gathered}$$

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