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Question Number 2748 by Rasheed Soomro last updated on 26/Nov/15

$$Showthat\frac{3^{2n+1}+5^{2n+1}}{8}isanintegerforn\in {\mathbb{Z}}^{+}.$$

Answered by prakash jain last updated on 26/Nov/15

$$f\left(n\right)=3^{2n+1}+5^{2n+1}$$$$f\left(1\right)=27+125=152=19\times 8$$$$\mathrm{Let}\mathrm{us}\mathrm{say}f\left(n\right)3^{2n+1}+5^{2n+1}\mathrm{is}\mathrm{divisible}\mathrm{by}8.$$$$f(n+1)=3^{2n+3}+5^{2n+3}$$$$f(\mathrm{n}+1)-f\left(n\right)=3^{2n+3}-3^{2n+1}+5^{2n+3}-5^{2n+1}$$$$=3^{2n+1}\left(8\right)+5^{2n+1}\left(24\right)$$$$=8(3^{2n+1}+3\cdot 5^{2\mathrm{n}+1})$$$$\because \frac{f\left(n\right)}{8}=k\Rightarrow f\left(n\right)=8k,k\in \mathbb{N}$$$$f(n+1)=8(3^{2n+1}+3\cdot 5^{2n+1})+8k$$$$f(n+1)=8(3^{2n+1}+3\cdot 5^{2n+1}+k)$$$$\therefore f(n+1)\mathrm{is}\mathrm{divisible}\mathrm{by}8\mathrm{or}$$$$\frac{f(n+1)}{8}\mathrm{is}\mathrm{an}\mathrm{integer}.$$

Commented by RasheedAhmad last updated on 26/Nov/15

$$\mathcal{N}ice!$$

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