Question

Tính ; A =1+\(\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2011}\left(1+2+3+...+2011\right)\)

Answer 1

Tổng các số tự nhiên từ 1 đến n là \(\frac{n\left(n+1\right)}{2}\)

Do đó \(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2011}.\frac{2011.2012}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2012}{2}\)

\(=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2012}{2}\right)-\frac{1}{2}\)

\(=\frac{1+2+3+...+2012}{2}-\frac{1}{2}\)

\(=\frac{\frac{2012.2013}{2}}{2}-\frac{1}{2}\)

\(=1012538,5\)

Vậy ....

Answer 2

A=(n+1)(n+2)/4=2012.2013/4=503.2013