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Question Number 66018 by Rio Michael last updated on 07/Aug/19

$$provebymathematicalinductionthat$$$$\underset{r=1}{\sum ^n}r(r+1)=\frac{n}{3}(n+1)(n+2)$$

Commented by Prithwish sen last updated on 07/Aug/19

$$\mathrm{when}\mathrm{n}=1$$$$1(1+1)=\frac{1}{3}(1+1)(1+2)$$$$2=2\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=1$$$$\mathrm{for}\mathrm{n}=2$$$$1(1+1)+2(2+1)=\frac{2}{3}(2+1)(2+2)$$$$8=8\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=2$$$$\mathrm{Now}\mathrm{let}\mathrm{us}\mathrm{assume}\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k}$$$$\mathrm{i}.\mathrm{e}$$$$\underset{\mathrm{r}=1}{\sum ^{\mathrm{k}}}\mathrm{r}(\mathrm{r}+1)=\frac{\mathrm{k}}{3}(\mathrm{k}+1)(\mathrm{k}+2)$$$$\Rightarrow 1(1+1)+2(2+1)+...\mathrm{k}(\mathrm{k}+1)=\frac{\mathrm{k}}{3}(\mathrm{k}+1)(\mathrm{k}+2)$$$$\mathrm{now}\mathrm{for}\mathrm{n}=\mathrm{k}+1$$$$\underset{\mathrm{r}=1}{\sum ^{\mathrm{k}+1}}\mathrm{r}(\mathrm{r}+1)=1(1+1)+2(2+1)+....+\mathrm{k}(\mathrm{k}+1)+(\mathrm{k}+1)(\mathrm{k}+2)$$$$=\frac{\mathrm{k}}{3}(\mathrm{k}+1)(\mathrm{k}+2)+(\mathrm{k}+1)(\mathrm{k}+2)=(\mathrm{k}+1)(\mathrm{k}+2)(\frac{\mathrm{k}}{3}+1)$$$$=\frac{(\mathrm{k}+1)}{3}[(\mathrm{k}+1)+1][(\mathrm{k}+1)+2]$$$$\therefore \mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k}+1\mathrm{also}.$$$$\mathrm{Hence}\mathrm{proved}.$$$$$$$$$$$$$$

Answered by Tanmay chaudhury last updated on 07/Aug/19

$$1.2+2.3+3.4+...+n(n+1)=\frac{n(n+1)(n+2)}{3}$$$$n=1$$$$LHS=1\times 2=2$$$$RHS=\frac{1(1+1)(1+2)}{3}=2$$$$LHS=RHS$$$$n=2$$$$LHS=1\times 2+2\times 3=8$$$$RHS=\frac{2(2+1)(2+2)}{3}=8$$$$LHS=RHS$$$$nowassumewhenn=kgivenstatementistrue$$$$wehaveprovethatitistruen=k+1$$$$1.2+2.3+3.4+...+k(k+1)=\frac{k(k+1)(k+2)}{3}$$$$nowLHS$$$$1.2+2.3+3.4+..+k(k+1)+(k+1)(k+2)$$$$=\frac{k(k+1)(k+2)}{3}+(k+1)\left(k\right)2)$$$$=(k+1)(k+2)(\frac{k}{3}+1)$$$$=\frac{(k+1)(k+2)(k+3)}{3}$$$$RHS$$$$=\frac{k(k+1)(k+2}{3}+(k+1)(k+2)$$$$=\frac{(k+1)(k+2)}{1}\times (\frac{k}{3}+1)$$$$=\frac{(k+1)(k+2)(k+3)}{3}$$$$LHS=RHS[henceproved$$

Commented by Rio Michael last updated on 07/Aug/19

$$thank{y}^{\prime }all$$

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