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KVPY Sample Paper KVPY Stream-SX Model Paper-4

  • question_answer
    If the sum \[\sum\limits_{k=1}^{\infty }{\,\frac{1}{(k+2)\sqrt{k}+\sqrt{k+2}}=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{c}}}\] \[a,b,c\text{ }\in \text{ }N\] and lie in [1, 15], then \[a\text{ }+\text{ }b\text{ }+\text{ }c\] equals to-

    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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