Question

Tính

+

1+(1+2)+(1+2+3)+...+(1+2+3+...+98) phần 1.2+2.3+3.4+...+98.99

Giúp vỚi mình cần gấp

Answer 1

Đặt \(A=1+\left(1+2\right)+\left(1+2+3\right)+\cdots+\left(1+2+3+\cdots+98\right)\)

\(=\frac{1\cdot2}{2}+\frac{2\cdot3}{2}+\frac{3\cdot4}{2}+\cdots+\frac{98\cdot99}{2}\)

\(=\frac{1\cdot2+2\cdot3+\cdots+98\cdot99}{2}\)

Ta có: \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+\cdots+\left(1+2+3+\cdots+98\right)}{1\cdot2+2\cdot3+\cdots+98\cdot99}\)

\(=\frac{\frac{1\cdot2+2\cdot3+\cdots+98\cdot99}{2}}{1\cdot2+2\cdot3+\cdots+98\cdot99}=\frac12\)