Question

tính tổng sau: A=1/2+1/2^2+1/2^3 +...+1/2^2015+1/2^2016

Answer 1

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)16 

2A=\(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}+\frac{1}{2017}\)

2A-A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)-\(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\)

A=\(\frac{1}{2017}-\frac{1}{2}\)

Answer 2

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

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

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

A = \(1-\frac{1}{2^{2016}}\)

Answer 3

A=\(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+......+\frac{1}{2^{2016}}\)

\(\frac{1}{2}\)A=\(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+.....+\frac{1}{2^{2017}}\)

Trừ vế cho vế ta có :\(\frac{1}{2^1}A=\frac{1}{2^1}-\frac{1}{2^{2017}}\)

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