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Question Number 162414 by HongKing last updated on 29/Dec/21

$$\mathrm{Prove}\mathrm{the}\mathrm{Identity}\mathrm{for}\mathrm{any}(\mathrm{a},\mathrm{n})\mathrm{in}\mathrm{Real}\mathrm{Number}$$$$(1+\mathrm{a})\cdot {\mathrm{a}}^{\left[\mathbf{n}\right]}=\mathrm{a}\cdot {\mathrm{a}}^{2\left[\frac{\mathbf{n}}{2}\right]}+{\mathrm{a}}^{2\left[\frac{\mathbf{n}+1}{2}\right]}$$$$[\ast ]\mathrm{Greatest}\mathrm{Integer}\mathrm{Function}$$

Answered by mindispower last updated on 31/Dec/21

$$n\in [2k,2k+1[$$$$\left[n\right]=2k$$$$(1+a).a^{2k}=a.a^{2k}+a^{2k}=(1+a).a^{2k}$$$$true$$$$n\in [2k-1,2k]$$$$\left[n\right]=2k-1$$$$\left[\frac{n}{2}\right]=k-1$$$$\left[\frac{n+1}{2}\right]=k$$$$(1+a).a^{2k-1}=?a.a^{2(k-1)}+a^{2k}=a^{2k-1}+a^{2k}=a^{2k-1}(1+a)$$$$True$$$$(1+a).a^{\left[n\right]}=a.a^{2\left[\frac{n}{2}\right]}+a^{2\left[\frac{n+1}{2}\right]}$$$$$$

Commented by HongKing last updated on 31/Dec/21

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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