A) \[-5050\]
B) \[0\]
C) \[5050\]
D) None of these
Correct Answer: C
Solution :
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{\left[ \sum\limits_{k=1}^{100}{{{x}^{k}}} \right]-100}{(x-1)}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{(x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{100}})-100}{(x-1)}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{(x-1)+({{x}^{2}}-1)+({{x}^{3}}-1)+...+({{x}^{100}}-1)}{(x-1)}\] \[=\lim \left\{ \left( \frac{x-1}{x-1} \right)+\left( \frac{{{x}^{2}}-1}{x-1} \right)+\left( \frac{{{x}^{3}}-1}{x-1} \right) \right.\] \[\left. +...+\left( \frac{{{x}^{100}}-1}{x-1} \right) \right\}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\left( \frac{x-1}{x-1} \right)+\underset{x\to 1}{\mathop{\lim }}\,\left( \frac{{{x}^{2}}-1}{x-1} \right)+\underset{x\to 1}{\mathop{\lim }}\,\left( \frac{{{x}^{3}}-1}{x-1} \right)\] \[+...+\underset{x\to 1}{\mathop{\lim }}\,\left( \frac{{{x}^{100}}-1}{x-1} \right)\] \[=1+2+3+...+100\] \[=\Sigma 100=\frac{100\times (100+1)}{2}\] \[=50\times 101\,=5050\] \[\left\{ \because \,\Sigma n=\frac{n(n+1)}{2} \right\}\].You need to login to perform this action.
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