Question

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

Tìm số nguyên dương x để B=115

Giải Nhanh Giups Mình Cần Gấp Lắm

 

Answer 1

Ta có :

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

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

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

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

để B = 115 thì \(\frac{2+3+4+...+\left(x+1\right)}{2}=115\)

\(\Rightarrow\)\(\left(x+3\right)x=115.2.2\)

\(\Rightarrow\)\(\left(x+3\right)x=23.20\)

\(\Rightarrow\)x = 20