Question

\(\lim\limits_{x\rightarrow1}\dfrac{x+x^2+...+x^n-n}{x-1}\)

Answer 1

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)+\left(x^2-1\right)+...+\left(x^n-1\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)+\left(x-1\right)\left(x+1\right)+...+\left(x-1\right)\left(x^{n-1}+x^{n-2}+...+1\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\left[1+\left(x+1\right)+\left(x^2+x+1\right)+...+\left(x^{n-1}+x^{n-2}+...+1\right)\right]\)

\(=1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)