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Question Number 73039 by mathmax by abdo last updated on 05/Nov/19

$$let{U}_n=\frac{n}{2}ifnevenand{U}_n=\frac{n-1}{2}ifnoddletf\left(n\right)=\sum _{k=0}^n{U}_k$$$$provethat\mathrm{\forall }(x,y)\in N^{2}f(x+y)-f(x-y)=xy$$

Answered by mind is power last updated on 05/Nov/19

$$\mathrm{just}\mathrm{for}\mathrm{the}\mathrm{definition}\mathrm{of}\mathrm{f}$$$$\mathrm{x}⩾\mathrm{y}$$$$\mathrm{x}+\mathrm{y},\mathrm{x}-\mathrm{y}\mathrm{has}\mathrm{sam}\mathrm{parite}$$$$\mathrm{f}\left(\mathrm{n}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{U}}_{\mathrm{k}}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{E}\left(\frac{\mathrm{n}}{2}\right)}}\mathrm{k}+\underset{\mathrm{k}=1}{\sum ^{\mathrm{E}\left(\frac{\mathrm{n}-1}{2}\right)}}\mathrm{k}=\frac{(\mathrm{E}\left(\frac{\mathrm{n}}{2}\right)+1).\left(\mathrm{E}\left(\frac{\mathrm{n}}{2}\right)\right)}{2}+\frac{\mathrm{E}\left(\frac{\mathrm{n}-1}{2}\right).(1+\mathrm{E}\left(\frac{\mathrm{n}-1}{2}\right))}{2}$$$$\mathrm{f}(\mathrm{x}+\mathrm{y})-\mathrm{f}(\mathrm{x}-\mathrm{y})$$$$\mathrm{is}\mathrm{x}+\mathrm{y}=2\mathrm{n}$$$$\mathrm{x}-\mathrm{y}=2\mathrm{n}-2\mathrm{y}$$$$\Rightarrow \mathrm{f}(\mathrm{x}+\mathrm{y})-\mathrm{f}(\mathrm{x}-\mathrm{y})=\frac{\mathrm{n}(\mathrm{n}+1)}{2}-\frac{(\mathrm{n}-\mathrm{y}+1)(\mathrm{n}-\mathrm{y})}{2}+\frac{(\mathrm{n}-1)\left(\mathrm{n}\right)}{2}-\frac{(\mathrm{n}-\mathrm{y}-1)(\mathrm{n}-\mathrm{y})}{2}$$$$=\frac{{\mathrm{n}}^2+\mathrm{n}-({\mathrm{n}}^2-2\mathrm{ny}+{\mathrm{y}}^2+\mathrm{n}-\mathrm{y})+{\mathrm{n}}^2-\mathrm{n}-({\mathrm{n}}^2+{\mathrm{y}}^2-2\mathrm{ny}-\mathrm{n}+\mathrm{y})}{2}$$$$=\frac{-{2\mathrm{y}}^2+4\mathrm{ny}}{2}=\frac{-{2\mathrm{y}}^2+2(\mathrm{x}+\mathrm{y})\mathrm{y}}{2}=\mathrm{xy}$$$$\mathrm{same}\mathrm{thing}\mathrm{if}\mathrm{x}+\mathrm{y}=2\mathrm{n}+1$$$$$$

Commented by mathmax by abdo last updated on 05/Nov/19

$$thanxsir.$$

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