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[Nguyen Ngoc Linh]

Tính tổng:

S= $\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}{2^{n}}$

[huykinhcan99]

Tính tổng:

$S=\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}{2^{n}}$

Ta có

\[\dfrac{2k-1}{2^k} = \dfrac{2\left(2k+1\right)-\left(2k+3\right)}{2^k} = \dfrac{2k+1}{2^{k-1}}-\dfrac{2k+3}{2^k}\]

 

Cho $k$ chạy từ $1$ đến $n$ rồi cộng lại, ta có

\[S=\dfrac{1}{2}+\dfrac{3}{2^2}+\ldots+\dfrac{2n-1}{2^n} = 3-\dfrac{2n+3}{2^n}\]

Edited by huykinhcan99, 21-11-2016 - 22:57.