Question

[Quỳnh Anh]

Rút gọn biểu thức \(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+\dfrac{3}{x^4}+...+\dfrac{n}{x^{n+1}}\) bằng:

A. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)

B. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{2n}\left(x-1\right)^2}\)

C. \(S=\dfrac{x^n-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)

D. \(S=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^n\left(x-1\right)^2}\)

Answer 1

[Nguyễn Việt Lâm]

\(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+...+\dfrac{n}{x^{n+1}}\)

\(\Rightarrow x.S\left(x\right)=\dfrac{1}{x}+\dfrac{2}{x^2}+\dfrac{3}{x^3}+...+\dfrac{n}{x^n}\)

\(\Rightarrow x.S\left(x\right)-S\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+...+\dfrac{1}{x^n}-\dfrac{n}{x^{n+1}}\)

\(\Rightarrow\left(x-1\right)S\left(x\right)=\dfrac{1}{x}.\dfrac{1-\left(\dfrac{1}{x}\right)^n}{1-\dfrac{1}{x}}-\dfrac{n}{x^{n+1}}=\dfrac{x^n-1}{x^n\left(x-1\right)}-\dfrac{n}{x^{n+1}}=\dfrac{x^{n+1}-x-n\left(x-1\right)}{x^{n+1}\left(x-1\right)}\)

\(\Rightarrow S\left(x\right)=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)