Question

Cho S=\(\dfrac{2}{2^1}\)+\(\dfrac{3}{2^2}\)+\(\dfrac{4}{2^3}\)+...+\(\dfrac{n+1}{2^n}\)+...+\(\dfrac{2017}{2^{2016}}\)

So sánh S với 3.

Answer 1

\(S=\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\)

\(\Rightarrow2S=2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\)

\(\Rightarrow2S-S=\left(2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\right)-\left(\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\right)\)

\(\Leftrightarrow S=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2017}{2^{2016}}\)

Tới đây thì đơn giản rồi nhé