A) \[\frac{9}{100}\]
B) \[-1/2\]
C) \[\frac{1}{99}\]
D) \[\frac{1}{101}\]
Correct Answer: B
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{{{1}^{99}}+{{2}^{99}}+.....+{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\,\sum\limits_{r=1}^{n}{\,\left( \frac{{{r}^{99}}}{{{n}^{100}}} \right)}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{1}{n}\,\,\sum\limits_{r=1}^{n}{\,{{\left( \frac{r}{n} \right)}^{99}}=\int_{0}^{1}{{{x}^{99}}dx=\left[ \frac{{{x}^{100}}}{100} \right]_{0}^{1}=\frac{1}{100}.}}\]
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