Tính tổng sau Sn=(2+1/2)^2+(4+1/4)^2+....+(2^n+1/2^n)^2

* Hướng dẫn giải

Chọn D

$\begin{array}{l}{\text{\hspace{0.33em}S}}_n=2^2+\frac{1}{2^2}+2+2^4+\frac{1}{2^4}+2+\mathrm{...}+2^{2n}+\frac{1}{2^{2n}}+2\\ =\left(2^2+2^4+\mathrm{...}+2^{2n}\right)+\left(\frac{1}{2^2}+\frac{1}{2^4}+\mathrm{...}+\frac{1}{2^{2n}}\right)+2n\\ =4.\frac{1-4^n}{1-4}+\frac{1}{4}\frac{1-\frac{1}{4^n}}{1-\frac{1}{4}}+2n\\ = \frac{4}{3}. (\hspace{0.17em}{4}^n-1)+\hspace{0.17em}\frac{1}{3}.\frac{4^n-1}{4^n}+\hspace{0.17em}2n \\ =\frac{4^n-1}{3}\left(4-\frac{1}{4^n}\right)+2n.\end{array}$