A) 6
B) 8
C) 10
D) 11
Correct Answer: D
Solution :
[D]Let \[{{T}_{k}}=\frac{(k+2)\,\sqrt{k}-k\,\sqrt{k+2}}{k\,\,{{(k+2)}^{2}}-{{k}^{2}}\,\,(k+2)}\]\[=\frac{(k+2)\,\,\sqrt{k}-k\,\sqrt{k+2}}{2k\,\,(k+2)}=\frac{1}{2}\left[ \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+2}} \right]\] |
\[\therefore {{T}_{1}}=\frac{1}{2}\left[ \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}} \right]\] |
\[{{T}_{2}}=\frac{1}{2}\left[ \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{4}} \right]\] \[{{T}_{3}}=\frac{1}{2}\left[ \frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}} \right]\] and so on \[\therefore As\,\,k\to \infty ,\] \[sum=\frac{1}{2}\left[ 1+\frac{1}{\sqrt{2}} \right]\] \[=\frac{1+\sqrt{2}}{2\sqrt{2}}=\frac{\sqrt{1}+\sqrt{2}}{\sqrt{8}}\] |
\[\Rightarrow a+b+c=11\] |
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