Find the sum $\frac2{\sqrt2+\sqrt6}+\frac2{\sqrt4+\sqrt8}+\frac2{\sqrt6+\sqrt{10}}+\ldots+\frac2{\sqrt{60}+\sqrt{64}}$

Question

If the sum of the series

$$\frac2{\sqrt2+\sqrt6}+\frac2{\sqrt4+\sqrt8}+\frac2{\sqrt6+\sqrt{10}}+\ldots+\frac2{\sqrt{60}+\sqrt{64}}=\sqrt{A}-\sqrt{B}+{C}$$

Find the value of $A+B$

Okay, this series is actually a tough one for me. I don't know at all how to approach this question. I would be really glad if someone will help me with this question

Answer 1

\begin{align}\sum_{i=1}^{30} \frac{2}{\sqrt{2i}+\sqrt{2i+4}}&=\sqrt{2} \sum_{i=1}^{30} \frac{1}{\sqrt{i}+\sqrt{i+2}}\\ &=\sqrt2 \sum_{i=1}^{30}\frac{\sqrt{i}-\sqrt{i+2}}{i-(i+2)} \\&=-\frac{\sqrt{2}}{2} \sum_{i=1}^{30}(\sqrt{i}-\sqrt{i+2}) \end{align}

You should be able to complete the problem using telescoping sum.