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

Here, general term ${\mathrm{T}}_{\mathrm{k}}=\frac{1}{(\mathrm{k}+2)\sqrt{\mathrm{k}}+\mathrm{k}\sqrt{\mathrm{k}+2}}$

Rationalising the expression, we get,

$\Rightarrow$ ${\text{T}}_{\text{k}}=\frac{\left(\text{k}+2\right)\sqrt{\text{k}}-\text{k}\sqrt{\text{k}+2}}{\text{k}{\left(\text{k}+2\right)}^2-{\text{k}}^2\left(\text{k}+2\right)}$

$\Rightarrow {\mathrm{T}}_{\mathrm{k}}=\frac{(\mathrm{k}+2)\sqrt{\mathrm{k}}-\mathrm{k}\sqrt{\mathrm{k}+2}}{2\mathrm{k}(\mathrm{k}+2)}=\frac{1}{2}(\frac{1}{\sqrt{\mathrm{k}}}-\frac{1}{\sqrt{\mathrm{k}+2}})$

Now, ${\mathrm{T}}_1=\frac{1}{2}(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}),$

${\mathrm{T}}_2=\frac{1}{2}(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{4}}),$

       ${\text{T}}_3=\frac{1}{2}\left[\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right]$  and so on

${\mathrm{T}}_{\mathrm{k}}=\frac{1}{2}(\frac{1}{\sqrt{\mathrm{k}}}-\frac{1}{\sqrt{\mathrm{k}+2}})$

Adding all the terms, we get,

${\mathrm{T}}_1+{\mathrm{T}}_2+...{\mathrm{T}}_{\mathrm{k}}=\frac{1}{2}(1+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{\mathrm{k}+1}}-\frac{1}{\sqrt{\mathrm{k}+2}})$

Now, if $\mathrm{k}\to \mathrm{\infty }$

$\sum _{\mathrm{k}=1}^{\mathrm{\infty }}{\mathrm{T}}_{\mathrm{k}}=\frac{1}{2}(1+\frac{1}{\sqrt{2}}-0-0)$

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

$\frac{\sqrt{1}+\sqrt{2}}{\sqrt{8}}=\frac{\sqrt{2}+\sqrt{4}}{\sqrt{16}}$

As, $\mathrm{c}\in [1,15]\Rightarrow \mathrm{c}=8, \mathrm{a}=1, \mathrm{b}=2$

$\Rightarrow$ any other solution is not possible

$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=11$