Question

Tính tổng sau: A = \(\frac{1}{2}\)+ \(\frac{1}{2^2}\) +  \(\frac{1}{2^3}\)+ ... + \(\frac{1}{2^{2015}}\)+ \(\frac{1}{2^{2016}}\)

Answer 1

Có : 

2A = 1+1/2+1/2^2+.....+1/2^2015

A = 2A - A = (1+1/2+1/2^2+......+1/2^2015) - (1/2+1/2^2+.....+1/2^2016)

   = 1 - 1/2^2016

Tk mk nha

Answer 2

Xét \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)

          \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)

Trừ vế theo vế ta được:

   \(2A-A=A=1-\frac{1}{2^{2016}}\)

Không thể tính được nữa vì số mũ quá lớn!!!

Answer 3

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\right)\)

\(A=1-\frac{1}{2^{2016}}\)