Giải thích các bước giải:
Ta có:
$\dfrac{k}{n(n+k)}=\dfrac{(n+k)-n}{n(n+k)}=\dfrac{n+k}{n(n+k)}-\dfrac{n}{n(n+k)}=\dfrac1n-\dfrac1{n+k}$
Áp dụng ta có:
$S=(\dfrac11-\dfrac13)+(\dfrac13-\dfrac15)+(\dfrac15-\dfrac17)+(\dfrac17-\dfrac19)+...+(\dfrac1{99}-\dfrac1{101})$
$\to S=\dfrac11-\dfrac13+\dfrac13-\dfrac15+\dfrac15-\dfrac17+\dfrac17-\dfrac19+...+\dfrac1{99}-\dfrac1{101}$
$\to S=\dfrac11-\dfrac1{101}$
$\to S=1-\dfrac1{101}$
$\to S=\dfrac{100}{101}$
