Question

[Beh5cyk]

B=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/x(1+2+3+....+x)

Giúp mình với mình đg cần gấp

Answer 1

[Ngô Văn Phương]

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+...+x\right)\)

\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}+\frac{4\cdot5}{2}+...+\frac{1}{x}\cdot\frac{x\left(x+1\right)}{2}\)

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

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

\(=\frac{1}{2}\cdot\frac{\left(x+1+2\right)\left(x+1-2+1\right)}{2}\)

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