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Question Number 27094 by abdo imad last updated on 02/Jan/18

$$if1+x+x^2=0findthevalueof$$$$A={(x+\frac{1}{x})}^6+{(x^2+\frac{1}{x^2})}^6+...{(x^{100}+\frac{1}{x^{100}})}^6.$$

Commented by AHSoomro last updated on 02/Jan/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$

Commented by abdo imad last updated on 08/Jan/18

$$A=\sum _{k=1}^{100}{(x^k+x^{-k})}^6=\sum _{k=1}^{k=100}(x^k+x_k^{-})buttherootsof$$$$1+x+x^2=0arej=e^{i\frac{2\pi }{3}}and{j}^{-}=e^{-i\frac{2\pi }{3}}wecanchoosex=e^{i\frac{2\pi }{3}}$$$$A=\sum _{k=1}^{100}{\left(2Re\left(j^k\right)\right)}^6=2^6\sum _{k=1}^{100}cos{\left(\frac{2k\pi }{3}\right)}^6$$$$=64\sum _{k=1}^{100}cos{\left(\frac{2k\pi }{3}\right)}^6$$$$but{\left(cosx\right)}^6={\left(\frac{e^{ix}+e^{-ix}}{2}\right)}^6=\frac{1}{2^6}\sum _{k=0}^6{C}_6^k{\left(e^{ix}\right)}^k{\left(e^{-ix}\right)}^{6-k}$$$$=\frac{1}{2^6}\sum _{k=0}^6{C}_6^k{e}^{ikx}{e}^{-i(6-k)x}...becontinued...$$$$$$

Answered by bsayani309@gmail.com last updated on 05/Jan/18

$$1+x+x^2=0$$$$or.1+x^2=-x$$$$or.\frac{1+x2}{x}=-1$$$$or.(\frac{1}{x}+x)=-1$$$$or.{(\frac{1}{x}+x)}^6=1$$$$thenA=1+2nd1+3rd1+.......+100th1=100$$

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