A) n
B) \[\frac{n+1}{2}\]
C) \[\frac{n(n+1)}{2}\]
D) \[\frac{n(n-1)}{2}\]
Correct Answer: C
Solution :
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{n}}-n}{x-1}\] \[\underset{x\to 1}{\mathop{\lim }}\,\frac{\left[ \begin{align} & (x-1)+({{x}^{2}}-{{1}^{2}})+({{x}^{3}}-{{1}^{3}})+ \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...+({{x}^{n}}-{{1}^{n}}) \\ \end{align} \right]}{x-1}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\left[ \frac{x-1}{x-1}+\frac{({{x}^{2}}-{{1}^{2}})}{x-1}+\frac{({{x}^{3}}-{{1}^{3}})}{x-1} \right.\] \[+...+\left. \frac{({{x}^{n}}-{{1}^{n}})}{x-1} \right]\] \[=1+2+3+4+...+n\]\[=\frac{n(n+1)}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
