Có 2 cách làm
Cách 1: Phân tích bình thường
\(x+x^2+...+x^n-n=\left(x-1\right)+\left(x^2-1\right)+...+\left(x^n-1\right)\)
\(=\left(x-1\right)+\left(x-1\right)\left(x+1\right)+\left(x-1\right)\left(x^2+x+1\right)+....+\left(x-1\right)\left(x^{n-1}+x^{n-2}+...+1\right)\)
\(=\left(x-1\right)\left(1+x+1+x^2+x+1+...+x^{n-1}+x^{n-2}+...+1\right)\)
\(=\left(x-1\right)\left(1.n+\left(n-1\right)x+\left(n-2\right)x^2+...+\left(n-n+2\right)x^{n-2}+\left(n-n+1\right)x^{n-1}\right)\)
\(=\left(x-1\right)\left(n+\left(n-1\right)x+\left(n-2\right)x^2+...+2x^{n-2}+x^{n-1}\right)\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{x+x^2+...+x^n-n}{x-1}=\dfrac{\left(x-1\right)\left[n+\left(n-1\right)x+\left(n-2\right)x^2+...+2x^{n-2}+x^{n-1}\right]}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left[n+\left(n-1\right)x+\left(n-2\right)x^2+...+2x^{n-2}+x^{n-1}\right]\)
\(=n+\left(n-1\right)+\left(n-2\right)+...+2+1=1+2+...+\left(n-1\right)+\left(n-2\right)=\dfrac{n\left(n+1\right)}{2}\)
Cách 2: Sử dụng L'Hospital
\(\lim\limits_{x\rightarrow1}\dfrac{x+x^2+...+x^n-n}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{1+2x+3x^2+...+nx^{n-1}}{1}=1+2.1+3.1+...+n=\dfrac{n\left(n+1\right)}{2}\)
