If the sum ∑k=1∞1(k+2)k+kk+2=1a−b where a,b∈N then a+b equals to

Tk=(k+2)k−kk+2k(k+2)2−k2(k+2)=(k+2)k−kk+22k(k+2)=12[1k−1k+2]T1=12[11−13], T2=12[12−14], T3=12[13−15]and so on ∴ as k→∞ sum =12[1+12]=1+222=1+28=116−8

If the sum ∑k=1∞1(k+2)k+kk+2=1a−b where a,b∈N then a+b equals to

Tk=(k+2)k−kk+2k(k+2)2−k2(k+2)=(k+2)k−kk+22k(k+2)=12[1k−1k+2]T1=12[11−13], T2=12[12−14], T3=12[13−15]and so on ∴ as k→∞ sum =12[1+12]=1+222=1+28=116−8

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If the sum $\sum_{k = 1}^{\infty} \frac{1}{(k + 2) \sqrt{k} + k \sqrt{k + 2}} = \frac{1}{\sqrt{a} - \sqrt{b}}$ where $a, b \in N$ then $a + b$ equals to

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Detailed Solution

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

$T_{1} = \frac{1}{2} [\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{3}}], T_{2} = \frac{1}{2} [\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{4}}], T_{3} = \frac{1}{2} [\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{5}}]$

and so on

$∴$ as $k \to \infty$ sum $= \frac{1}{2} [1 + \frac{1}{\sqrt{2}}] = \frac{1 + \sqrt{2}}{2 \sqrt{2}} = \frac{\sqrt{1} + \sqrt{2}}{\sqrt{8}} = \frac{1}{\sqrt{16} - \sqrt{8}}$