Tính $\frac{1}{3}+\frac{1}{3+6}+\frac{1}{3+6+9}+...+\frac{1}{3+6+9+...+2013}$

Question

Tính

$\frac{1}{3}+\frac{1}{3+6}+\frac{1}{3+6+9}+...+\frac{1}{3+6+9+...+2013}$

score 3 · Answer 1

$\frac{1}{3}+\frac{1}{3+6}+\frac{1}{3+6+9}+...+\frac{1}{3+6+9+...+2013}$

$=\frac{1}{3}\left\{\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+....+n}\right\}$ làm tổng quát với n sau đó áp dụng với

$n=671$

$=\frac{1}{3}\left\{\frac{2}{1(1+1)}+\frac{2}{2(2+1)}+\frac{2}{3(3+1)}+...+\frac{2}{n(n+1)}\right\}$

$=\frac{2}{3}\sum_{k=1}^{n}\frac{1}{k(k+1)}$

$=\frac{2}{3}\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)$

$=\frac{2}{3}\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{n+1}\right)$

$=\frac{2}{3}\left(1-\frac{1}{n+1}\right)$

áp dụng với n=671 ta được

$A= \frac{2}{3}\left(1-\frac{1}{671+1}\right)=\frac{2}{3}\frac{671}{672}=\frac{1342}{2016}$

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