Question

tính tổng

S = \(\frac{1}{2^{-2013}+1}\) + \(\frac{1}{2^{-2012}+1}\) +....+ \(\frac{1}{2^0+1}\)+...+ \(\frac{1}{2^{2012}+1}\) +\(\frac{1}{2^{2013}+1}\)

Answer 1

Xét với n là số nguyên thì : \(\frac{1}{2^{-n}+1}+\frac{1}{2^n+1}=\frac{1}{\frac{1}{2^n}+1}+\frac{1}{2^n+1}=\frac{2^n}{2^n+1}+\frac{1}{2^n+1}=\frac{2^n+1}{2^n+1}=1\)

Vậy ta nhóm hợp lí như sau : 

\(S=\left(\frac{1}{2^{-2013}+1}+\frac{1}{2^{2013}+1}\right)+\left(\frac{1}{2^{-2012}+1}+\frac{1}{2^{2012}+1}\right)+...+\left(\frac{1}{2^{-1}+1}+\frac{1}{2^1+1}\right)+\frac{1}{2^0+1}\)

\(=1+1+...+1+\frac{1}{2}\) (2013 số hạng 1)

\(=2013+\frac{1}{2}\)