Value of $\sum_{n=0}^{1947}\frac{1}{2^n+\sqrt{2^{1947}}}$ $?$ 
Find the value of $$\sum_{n=0}^{1947}\frac{1}{2^n+\sqrt{2^{1947}}}$$

MY APPROACH : I'm not able to telescope it . I tried to Rationalize it , but It was not possible . I've  not attached any solution because , nothing striked me .
Here's a solution from Wolfram Alpha
 A: Hint:
$$\frac1{x+\sqrt{xy}}+\frac1{y+\sqrt{xy}}=\frac1{\sqrt x+\sqrt y}\left(\frac1{\sqrt x}+\frac1{\sqrt y}\right)=\frac1{\sqrt{xy}}.$$
See if you can use this to group some of the terms in your sum.
A: Notice that $\displaystyle \frac{1}{2^r+2^{\frac{1947}{2}}}+ \frac{1}{2^{1947-r}+2^{\frac{1947}{2}}}= \frac{1}{2^\frac{1947}{2}}$
$974$ such pairs can be formed.
Hence, the answer is $\displaystyle\frac{974}{2^\frac{1947}{2}}$
A: Hint 1:

$$\sum_{n=0}^{1947}\frac{1}{2^n+\sqrt{2^{1947}}} = \frac12 \sum_{n=0}^{1947} \left(\frac{1}{2^n+\sqrt{2^{1947}}}+\frac{1}{2^{1947 - n}+\sqrt{2^{1947}}} \right)$$

Hint 2:

$$\frac12 \sum_{n=0}^{1947} \left(\frac{1}{2^n+\sqrt{2^{1947}}}+\frac{1}{2^{1947 - n}+\sqrt{2^{1947}}} \right) = \frac12\cdot \frac{1}{\sqrt{2^{1947}}} $$

Hope it helps.
A: We will use the property that $$\sum_{r=a}^{b}f(r)=\sum_{r=a}^{b}f(a+b-r)$$
Therefore  $$S=\sum_{n=0}^{1947}\frac{1}{2^n+\sqrt{2^{1947}}}= \sum_{n=0}^{1947}\frac{1}{2^{1947-n}+\sqrt{2^{1947}}}$$
Let $k=1947$
$$S=\sum_{n=0}^{k}\frac{1}{2^n+\sqrt{2^{k}}}= \sum_{n=0}^{k}\frac{1}{2^{k-n}+\sqrt{2^{k}}}$$
Now We use second expression $$S=\sum_{n=0}^{k}\frac{1}{2^{k-n}+\sqrt{2^{k}}}$$
$$S=\sum_{n=0}^{k}\frac{2^n}{2^{k}+2^n\sqrt{2^{k}}}$$
$$S=\frac{1}{\sqrt{2^k}}\sum_{n=0}^{k}\frac{2^n}{\sqrt{2^{k}}+2^n}$$
$$S=\frac{1}{\sqrt{2^k}}\bigg(\sum_{n=0}^{k}\frac{2^n+\sqrt{2^k}}{\sqrt{2^{k}}+2^n}-\sum_{n=0}^{k}\frac{\sqrt{2^k}}{\sqrt{2^{k}}+2^n}\bigg)$$
$$S=\frac{1}{\sqrt{2^k}}((k+1)-\sqrt{2^k}S)$$
$$2S=\frac{k+1}{\sqrt{2^k}}$$
$$2S=\frac{1948}{\sqrt{2^1947}}$$
$$S=\boxed{\frac{974}{\sqrt{2^{1947}}}}$$
A: Here $$T_n=\frac{1}{2^n+\sqrt{2^{1947}}}$$
For simplicity let $\color{green}{x=\sqrt{2^{1947}}}$ . Now $$T_{1947-n}=\frac{1}{2^{1947-n}+x}\Rightarrow\frac{2^n}{x^2+2^nx}$$
Notice $\color{red}{T_n+T_{1947-n}}$ $$T_n+T_{1947-n}=\frac{1}{2^n+x}+\frac{2^n}{x^2+2^nx}$$ $$=\frac{1}{2^n+x}\left(1+\frac{2^n}{x}\right)\Rightarrow\color{red}{\frac{1}{x}}$$
Now you can simply compute the sum.
