Tính tổng $S=\sum_{k=1}^{n}\frac{k+2}{k(k+1)2^{k}}$
$\sum_{k=1}^n\frac{k+2}{k(k+1)2^k}= \sum_{k=1}^n\frac{(2k+2)-k}{k(k+1)2^k}= \sum_{k=1}^n\left(\frac{1}{k2^{k-1}}-\frac{1}{(k+1)2^k}\right)= \sum_{k=1}^n-\Delta\left(\frac{1}{k2^{k-1}}\right) $
$=-\left(\frac{1}{(n+1)2^n}-\frac{1}{1.2^{1-1}}\right)$
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P/s: Khảo cổ
