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

Here, general term $T_{k}=\frac{1}{\left(k + 2\right) \sqrt{k} + k \sqrt{k + 2}}$
Rationalising the expression, we get,
$\Rightarrow $ $T_{k}=\frac{\left(k + 2\right) \sqrt{k} - k \sqrt{k + 2}}{k \left(k + 2\right)^{2} - k^{2} \left(k + 2\right)}$
$\Rightarrow T_{k}=\frac{\left(k + 2\right) \sqrt{k} - k \sqrt{k + 2}}{2 k \left(k + 2\right)}=\frac{1}{2}\left(\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 2}}\right)$
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 $k \rightarrow \in fty$
$\displaystyle \sum _{k = 1}^{\in fty}T_{k}=\frac{1}{2}\left(1 + \frac{1}{\sqrt{2}} - 0 - 0\right)$
$=\frac{\sqrt{1} + \sqrt{2}}{2 \sqrt{2}}$
$\frac{\sqrt{1} + \sqrt{2}}{\sqrt{8}}=\frac{\sqrt{2} + \sqrt{4}}{\sqrt{16}}$
As, $c\in \left[1 , 15\right]\Rightarrow c=8,a=1,b=2$
$\Rightarrow $ any other solution is not possible
$\Rightarrow a+b+c=11$