Question

[Lại Đình Văn]

Tính tổng: \(A=\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+\frac{1}{4.\left(1+2+3+4\right)}+...+\frac{1}{2013.\left(1+2+3+...+2013\right)}\)

Answer 1

[Akai Haruma]

Lời giải:

** Sửa đề:

$A=\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+\frac{1}{4}(1+2+3+4)+....+\frac{1}{2013}(1+2+3+...+2013)$

$A=\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+....+\frac{1}{2013}.\frac{2013.2014}{2}$

$=\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{2014}{2}$

$=\frac{3+4+5+...+2014}{2}$

$=\frac{1+2+3+4+5+...+2014}{2}-\frac{3}{2}$
$=\frac{2014.2015:2}{2}-\frac{3}{2}$

$=1014551$