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

Question

$$\sum_{i=0}^{n} 2^{i/2} = (1+ \sqrt2)\left(2^{\frac{n+1}2} -1\right)$$

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.

Answer 1

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

Answer 2

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}