Question

tính:

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

Answer 1

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

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2016}}\right)\)

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

Vậy....
 

Answer 2

đặt dãy số là A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)

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

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

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

=>A=1-1/2²⁰¹⁶

A=.....