Question

Tính giá trị của biểu thức :

S=\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+.....+\frac{1}{100}.\left(1+2+...+100\right)\)

Ta được S={..}

Answer 1

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.....+\frac{1}{100}\left(1+2+3+....+100\right)\)

\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+\frac{1}{4}.\frac{4\left(4+1\right)}{2}+.....+\frac{1}{100}.\frac{100\left(100+1\right)}{2}\)

\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{100+1}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{101}{2}\)

\(=\frac{2+3+4+....+101}{2}\)

\(=\frac{\frac{101\left(101+1\right)}{2}-1}{2}=5150.5\)