Question Number 162414 by HongKing last updated on 29/Dec/21
![Prove the Identity for any (a,n) in Real Number (1 + a)∙a^([n]) = a ∙ a^(2[(n/2)]) + a^(2[((n+1)/2)]) [∗] Greatest Integer Function](https://www.tinkutara.com/question/Q162414.png)
$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{Identity}\:\mathrm{for}\:\mathrm{any}\:\left(\mathrm{a},\mathrm{n}\right)\:\mathrm{in}\:\mathrm{Real}\:\mathrm{Number} \\ $$$$\left(\mathrm{1}\:+\:\mathrm{a}\right)\centerdot\mathrm{a}^{\left[\boldsymbol{\mathrm{n}}\right]} \:=\:\mathrm{a}\:\centerdot\:\mathrm{a}^{\mathrm{2}\left[\frac{\boldsymbol{\mathrm{n}}}{\mathrm{2}}\right]} \:+\:\mathrm{a}^{\mathrm{2}\left[\frac{\boldsymbol{\mathrm{n}}+\mathrm{1}}{\mathrm{2}}\right]} \\ $$$$\left[\ast\right]\:\mathrm{Greatest}\:\mathrm{Integer}\:\mathrm{Function} \\ $$
Answered by mindispower last updated on 31/Dec/21
![n∈[2k,2k+1[ [n]=2k (1+a).a^(2k) =a.a^(2k) +a^(2k) =(1+a).a^(2k) true n∈[2k−1,2k] [n]=2k−1 [(n/2)]=k−1 [((n+1)/2)]=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^([n]) =a.a^(2[(n/2)]) +a^(2[((n+1)/2)])](https://www.tinkutara.com/question/Q162689.png)
$${n}\in\left[\mathrm{2}{k},\mathrm{2}{k}+\mathrm{1}\left[\right.\right. \\ $$$$\left[{n}\right]=\mathrm{2}{k} \\ $$$$\left(\mathrm{1}+{a}\right).{a}^{\mathrm{2}{k}} ={a}.{a}^{\mathrm{2}{k}} +{a}^{\mathrm{2}{k}} =\left(\mathrm{1}+{a}\right).{a}^{\mathrm{2}{k}} \\ $$$${true} \\ $$$${n}\in\left[\mathrm{2}{k}−\mathrm{1},\mathrm{2}{k}\right] \\ $$$$\left[{n}\right]=\mathrm{2}{k}−\mathrm{1} \\ $$$$\left[\frac{{n}}{\mathrm{2}}\right]={k}−\mathrm{1} \\ $$$$\left[\frac{{n}+\mathrm{1}}{\mathrm{2}}\right]={k} \\ $$$$\left(\mathrm{1}+{a}\right).{a}^{\mathrm{2}{k}−\mathrm{1}} =?\:{a}.{a}^{\mathrm{2}\left({k}−\mathrm{1}\right)} +{a}^{\mathrm{2}{k}} ={a}^{\mathrm{2}{k}−\mathrm{1}} +{a}^{\mathrm{2}{k}} ={a}^{\mathrm{2}{k}−\mathrm{1}} \left(\mathrm{1}+{a}\right) \\ $$$${True} \\ $$$$\left(\mathrm{1}+{a}\right).{a}^{\left[{n}\right]} ={a}.{a}^{\mathrm{2}\left[\frac{{n}}{\mathrm{2}}\right]} +{a}^{\mathrm{2}\left[\frac{{n}+\mathrm{1}}{\mathrm{2}}\right]} \\ $$$$ \\ $$
Commented by HongKing last updated on 31/Dec/21
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$