A) \[\frac{1}{2}\]
B) \[1\]
C) \[-1\]
D) \[-\frac{1}{2}\]
Correct Answer: D
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\,\,\left\{ \frac{1+2+3+....+n}{n+2}-\frac{n}{2} \right\}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\left\{ \frac{n(n+1)}{2}\times \frac{1}{n+2}-\frac{n}{2} \right\}\] \[\left\{ \because \,\,\,\sum\limits_{r=1}^{n}{r=\frac{n(n+1)}{2}} \right\}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\left\{ \frac{(n+1)}{2(n+2)}-\frac{1}{2} \right\}.n\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{2}\left\{ \frac{n+1-n-2}{n+2} \right\}.n\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{n}{2}\times \frac{(-1)}{n+2}=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{-1}{2} \right).\frac{1}{\left( 1+\frac{2}{n} \right)}\] \[=-\frac{1}{2}\times 1=-\frac{1}{2}\]
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