A) \[2+2\sqrt{2}\]
B) \[2\sqrt{2}-2\]
C) \[2\sqrt{2}\]
D) 2
Correct Answer: B
Solution :
\[y=\underset{n\to \infty }{\mathop{\lim }}\,\,\left[ \frac{1}{n}+\frac{1}{\sqrt{{{n}^{2}}+n}}+....+\frac{1}{\sqrt{{{n}^{2}}+(n-1)n}} \right]\] Þ \[y=\underset{n\to \infty }{\mathop{\lim }}\,\,\left[ \frac{1}{n}+\frac{1}{n\sqrt{1+\frac{1}{n}}}+....+\frac{1}{n\sqrt{1+\frac{(n-1)}{n}}} \right]\] Þ \[y=\frac{1}{n}\underset{n\to \infty }{\mathop{\lim }}\,\,\left[ 1+\frac{1}{\sqrt{1+\frac{1}{n}}}+....+\frac{1}{\sqrt{1+\frac{(n-1)}{n}}} \right]\] \[y=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{1+\frac{(k-1)}{n}}}}\], Put \[\frac{k-1}{n}=x\] and \[\frac{1}{n}=dx\] Þ \[y=\underset{n\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{\frac{n-1}{n}}{\frac{dx}{\sqrt{1+x}}}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,2\left[ \sqrt{1+x} \right]_{\,0}^{\,\left( \frac{n-1}{n} \right)}\] Þ \[y=2\underset{n\to \infty }{\mathop{\lim }}\,\,\left[ \sqrt{\frac{2n-1}{n}}-1 \right]\] \[=2\underset{n\to \infty }{\mathop{\lim }}\,\sqrt{\frac{2n-1}{n}}-2\] Þ \[y=2\underset{n\to \infty }{\mathop{\lim }}\,\sqrt{2-\frac{1}{n}}-2\]\[=2\sqrt{2}-2\].
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