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Question Number 34094 by Rio Mike last updated on 30/Apr/18

Commented by math khazana by abdo last updated on 01/May/18

$${(x^2+\frac{1}{x^2})}^8=\sum _{k=0}^8{C}_8^k{\left(x^2\right)}^k{\left(\frac{1}{x^2}\right)}^{8-k}$$$$=\sum _{k=0}^8{C}_8^k{x}^{2k}{x}^{2k-16}$$$$\sum _{k=0}^8{C}_8^k{x}^{4k-16}theconstanttermisobtained$$$$when4k-16=0\Rightarrow k=4andthecoefficientis$$$$C_8^4=\frac{8!}{4!.4!}=\frac{8.7.6.5}{4.3.2}=\frac{8.7.5}{4}=2.7.5=70.$$$$$$

Commented by math khazana by abdo last updated on 01/May/18

$$\sum _{r=1}^{\mathrm{\infty }}{3}^{2-r}=9\sum _{r=1}^{\mathrm{\infty }}{\left(\frac{1}{3}\right)}^r$$$$=9(\sum _{r=0}^{\mathrm{\infty }}{\left(\frac{1}{3}\right)}^r-1)$$$$=9\frac{1}{1-\frac{1}{3}}-9=9\frac{1}{\frac{2}{3}}-9=9.\frac{3}{2}-9$$$$=\frac{27-18}{2}=\frac{9}{2}.$$

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