Question

cho B=1+\(\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{x}\left(1+2+3+...+x\right)\)

Answer 1

\(B=1+\dfrac{1}{2}\left(\dfrac{2.3}{2}\right)+\dfrac{1}{3}\left(\dfrac{3.4}{2}\right)+\dfrac{1}{4}.\left(\dfrac{4.5}{2}\right)+...+\dfrac{1}{x}.\left(\dfrac{x\left(x+1\right)}{2}\right)\)

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

Từ đó \(B=115\) khi \(\dfrac{1}{2}.\left(\dfrac{x\left(x+3\right)}{2}\right)=115\Leftrightarrow x\left(x+3\right)=460\)

Mà \(x\) là số nguyên dương nên \(x\) và \(x+3\) là ước dương của 460 nên \(x=20\)

Vậy \(x=20\)