Question Number 34094 by Rio Mike last updated on 30/Apr/18
Commented by math khazana by abdo last updated on 01/May/18
$$\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{8}} \:=\:\sum_{{k}=\mathrm{0}} ^{\mathrm{8}} \:\:{C}_{\mathrm{8}} ^{{k}} \:\left({x}^{\mathrm{2}} \right)^{{k}} \:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{8}−{k}} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{\mathrm{8}} \:{C}_{\mathrm{8}} ^{{k}} \:{x}^{\mathrm{2}{k}} \:\:{x}^{\mathrm{2}{k}\:−\mathrm{16}} \:\:\:\: \\ $$$$\sum_{{k}=\mathrm{0}} ^{\mathrm{8}} \:\:{C}_{\mathrm{8}} ^{{k}} \:\:\:\:{x}^{\mathrm{4}{k}−\mathrm{16}} \:\:\:{the}\:{constant}\:{term}\:{is}\:{obtained} \\ $$$${when}\:\mathrm{4}{k}−\mathrm{16}\:=\mathrm{0}\:\Rightarrow{k}=\mathrm{4}\:\:{and}\:{the}\:{coefficient}\:{is} \\ $$$${C}_{\mathrm{8}} ^{\mathrm{4}} \:\:\:=\frac{\mathrm{8}!}{\mathrm{4}!\:.\mathrm{4}!}\:=\frac{\mathrm{8}.\mathrm{7}.\mathrm{6}.\mathrm{5}}{\mathrm{4}.\mathrm{3}.\mathrm{2}}\:=\frac{\mathrm{8}.\mathrm{7}.\mathrm{5}}{\mathrm{4}}\:=\:\mathrm{2}.\mathrm{7}.\mathrm{5}=\:\mathrm{70}. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 01/May/18
$$\sum_{{r}=\mathrm{1}} ^{\infty} \:\mathrm{3}^{\mathrm{2}−{r}\:} \:=\mathrm{9}\:\sum_{{r}=\mathrm{1}} ^{\infty} \:\:\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{r}} \\ $$$$=\mathrm{9}\left(\:\:\:\sum_{{r}=\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{r}} \:−\mathrm{1}\right) \\ $$$$=\mathrm{9}\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}\:−\mathrm{9}\:=\:\mathrm{9}\:\frac{\mathrm{1}}{\frac{\mathrm{2}}{\mathrm{3}}}\:\:−\mathrm{9}\:=\:\mathrm{9}.\frac{\mathrm{3}}{\mathrm{2}}\:−\mathrm{9} \\ $$$$=\:\:\frac{\mathrm{27}\:−\mathrm{18}}{\mathrm{2}}\:\:=\:\frac{\mathrm{9}}{\mathrm{2}}\:. \\ $$