Question

Tính \(A=2013+\frac{2013}{1+2}+\frac{2013}{1+2+3}+\frac{2013}{1+2+3+4}+...+\frac{2013}{1+2+3+...+2012}\)

Answer 1

Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)

Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)

\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)

\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)

\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)

\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)