Question

If the sum ${\sum }_{k-1}^{\infty }\frac{1}{(k+2)\sqrt{k}+k\sqrt{k+2}}=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{c}}$ where $a,b,cϵN$ and lie in [1, 15], then a+b+c equals to

• A. 6
• B. 8
• C. 10
• D. 11

Answer 1

The correct option is D

11

Let $T_k=\frac{(k+2)\sqrt{k}-k\sqrt{k+2}}{k(k+2{)}^2-k^2(k+2)}$

$=\frac{(k+2)\sqrt{k}-k\sqrt{k+2}}{2k(k+2)}=\frac{1}{2}[\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+2}}]$

$\therefore T_1=\frac{1}{2}[\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}]$

$T_2=\frac{1}{2}[\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{4}}]$

$T_3=\frac{1}{2}[\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}]$ and so on

$\therefore$ As $k\to \infty$, sum = $\frac{1}{2}[1+\frac{1}{\sqrt{2}}]=\frac{1+\sqrt{2}}{2\sqrt{2}}$

$=\frac{\sqrt{1}+\sqrt{2}}{\sqrt{8}}$

$\Rightarrow a+b+c=11$