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Question Number 103700 by anonymous last updated on 16/Jul/20

Commented by anonymous last updated on 16/Jul/20

$$plzzhelpme$$

Answered by bobhans last updated on 17/Jul/20

$${(x+\frac{2}{x})}^8=\frac{{(x^2+2)}^8}{x^8}=\frac{\underset{n=0}{\sum ^8}C{}_n^8{x}^{2n}.2^{8-n}}{x^8}$$$$=\frac{2^8+8\times 2^7{x}^2+28\times 2^6{x}^4+56\times 2^5{x}^6+...}{x^8}$$$$=\frac{256}{x^8}+\frac{1024}{x^6}+\frac{28\times 64}{x^4}+\frac{56+32}{x^2}+...$$

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