Đáp án:
Giải thích các bước giải:
$3 + \dfrac{3}{1+2+3} + \dfrac{3}{1+2+3+4} + ... + \dfrac{3}{1+2+...+100 }$
$⇒ 3 ( 1+ \dfrac{1}{1+2+3} \dfrac{1}{1+2+3+4} + ... + \dfrac{1}{1+2+..+100} )$
$⇒ 3 ( 1 + \dfrac{1}{2.3} \dfrac{1}{3.4} + ... + \dfrac{1}{100.101} )$
$⇒ 3 ( 1 + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + .... + \dfrac{1}{100} - \dfrac{1}{101} )
$⇒ 3 ( 1 + \dfrac{1}{2} - \dfrac{1}{101} )$
$⇒ 3 . \dfrac{301}{202}$
$⇒ \dfrac{903}{202}$