# How to find the value of $\sum_{n=0}^{1947} \left(\frac 1{2^n+ \sqrt{2^{1947}}} \right)$

I've no clue where to start from. I tried writing down a few terms to find out a pattern but didn't notice any. Any help would be appreciated! :)

• $$\frac{1}{2^n + \sqrt{2^{1947}}} + \frac{1}{2^{1947-n} + \sqrt{2^{1947}}} =\;?$$ – achille hui Nov 2 '14 at 15:59
• $$\frac{1}{2^n + \sqrt{2^{1947}}} + \frac{1}{2^{1947-n} + \sqrt{2^{1947}}} = \frac{\sqrt{2^{1947}}}{\sqrt{2^{1947}}(2^n + \sqrt{2^{1947}})} + \frac{2^n}{\sqrt{2^{1947}}^2 + 2^n\sqrt{2^{1947}}}\\ = \frac{\sqrt{2^{1947}}}{\sqrt{2^{1947}}(2^n + \sqrt{2^{1947}})} + \frac{2^n}{\sqrt{2^{1947}}(\sqrt{2^{1947}} + 2^n)} = \frac{1}{\sqrt{2^{1947}}}$$ Is that obvious now? – achille hui Nov 2 '14 at 16:11