Question

If ${\Sigma }_{k=1}^{\infty }\frac{1}{\left(k+2\right)\sqrt{k}+k\sqrt{k+2}}=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{c}}$ , where $a,b,c\in N$ and $a,b,c\in \left[1,15\right]$ , then $a+b+c$ is equal to

Answer 1

Answer: 11

Here, general term $T_k=\frac{1}{\left(k+2\right)\sqrt{k}+k\sqrt{k+2}}$
Rationalising the expression, we get,
$\Rightarrow$ $T_k=\frac{\left(k+2\right)\sqrt{k}-k\sqrt{k+2}}{k{\left(k+2\right)}^2-k^2\left(k+2\right)}$
$\Rightarrow T_k=\frac{\left(k+2\right)\sqrt{k}-k\sqrt{k+2}}{2k\left(k+2\right)}=\frac{1}{2}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+2}}\right)$
Now, $T_1=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}\right),$
$T_2=\frac{1}{2}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{4}}\right),$
$T_3=\frac{1}{2}\left[\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right]$ and so on
$T_k=\frac{1}{2}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+2}}\right)$
Adding all the terms, we get,
$T_1+T_2+...T_k=\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{k+1}}-\frac{1}{\sqrt{k+2}}\right)$
Now, if $k\to \in fty$
$\sum _{k=1}^{\in fty}{T}_k=\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}-0-0\right)$
$=\frac{\sqrt{1}+\sqrt{2}}{2\sqrt{2}}$
$\frac{\sqrt{1}+\sqrt{2}}{\sqrt{8}}=\frac{\sqrt{2}+\sqrt{4}}{\sqrt{16}}$
As, $c\in \left[1,15\right]\Rightarrow c=8,a=1,b=2$
$\Rightarrow$ any other solution is not possible
$\Rightarrow a+b+c=11$