Đáp án: $ A=\dfrac9{11}$
Giải thích các bước giải:
Ta có:
$S=1+2+3+...+n$
$\to S=n+(n-1)+(n-2)+...+1$
$\to 2S=(n+1)+(n-1+2)+(n-2+3)+..+(n+1)$ có $n$ số hạng
$\to 2S=(n+1)+(n+1)+(n+1)+...+(n+1)$
$\to 2S=n(n+1)$
$\to S=\dfrac12n(n+1)$
$\to 1+2+3+...+n=\dfrac12n(n+1)$
$\to \dfrac1{1+2+3+..+n}=\dfrac{2}{n(n+1)}=2\cdot\dfrac{1}{n(n+1)}=2\cdot\dfrac{n+1-n}{n(n+1)}=2(\dfrac1n-\dfrac1{n+1})$
Áp dụng ta được :
$A=2(\dfrac12-\dfrac13)+2(\dfrac13-\dfrac14)+2(\dfrac14-\dfrac15)+..+2(\dfrac1{10}-\dfrac1{11})$
$\to A=2(\dfrac12-\dfrac13+\dfrac13-\dfrac14+\dfrac14-\dfrac15+...+\dfrac1{10}-\dfrac1{11}))$
$\to A=2\cdot (\dfrac12-\dfrac1{11})$
$\to A=\dfrac9{11}$
