Q:

Evaluating sum of $\sum_{i=0}^{n} 2^{i/2}$

$$\sum_{i=0}^{n} 2^{i/2} = (1+ \sqrt2)\left(2^{\frac{n+1}2} -1\right)$$
I know the above is true, but how would I get the right hand side? This summation shows up from a algorithm recurrence problem I have.  

A:

Hint:
$$\sum_{i=0}^n 2^{i/2}=\sum_{i=0}^n (\sqrt{2})^i,$$ 
i.e. a geometric series...

A:

As
$\sqrt 2 + 1 = \frac1 {\sqrt 2 - 1}
$
you should compute 
\begin{align}(\sqrt 2 - 1)\times
\sum_{i=0}^{n} 2^{i/2} &= \sum_{i=0}^{n} 2^{i/2}2^{1/2} - \sum_{i=0}^{n} 2^{i/2} 
\\&=  \sum_{i=1}^{n+1} 2^{i/2} - \sum_{i=0}^{n} 2^{i/2} 
\\&= 2^{(n+1)/2} - 1
\end{align}

