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Question Number 116685 by mathocean1 last updated on 05/Oct/20

$$\mathrm{Given}\mathrm{the}\mathrm{equality}:$$$$1+3+5+...+(2\mathrm{p}+1)={(\mathrm{p}+1)}^2\mathrm{p}\in {\mathbb{N}}^{\ast }$$$$$$$$\mathrm{Show}\mathrm{this}\mathrm{equality}\mathrm{is}\mathrm{true}\mathrm{when}$$$$\mathrm{we}\mathrm{replace}\mathrm{p}\mathrm{by}\mathrm{p}+1$$

Answered by TANMAY PANACEA last updated on 05/Oct/20

$$1+(n-1)2=2p+1$$$$2n-1=2p+1\to n=p+1$$$$sumof(p+1)terms$$$$\frac{p+1}{2}[2\times 1+(p+1-1)2]$$$$\frac{(p+1)}{2}\times [2p+2]={(\mathit{p}+1)}^2$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{s}\mathit{u}\mathit{m}\mathit{o}\mathit{f}(\mathit{p}+1+1)\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{s}\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathit{L}HS$$$$\frac{(\mathit{p}+2)}{2}[2\times 1+(\mathit{p}+2-1)\times 2]$$$$=\frac{(\mathit{p}+2)}{2}\times [2+2\mathit{p}+2]\to {(\mathit{p}+2)}^2$$$$RHS{(p+1+1)}^2\to {(\mathit{p}+2)}^2$$$$\mathit{h}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{d}$$

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