Summation of $2^n$ where $n$ is even or odd I need the summation formula for the following:
$2^0+2^1+2^3+2^5+\cdots+2^n$, $n$ is even i.e. 
$2^0+2^2+2^4+2^6+\cdots+2^n$, $n$ is odd
I am aware that the formula of
$2^0+2^1+2^2+2^3+\cdots+2^n = 2^{n+1}-1$. But can anyone help me with a approach so solve such kind of problems.
 A: Hint: $$\sum_{k=0}^m2^{2k} = \sum_{k=0}^m\left(2^2\right)^{k}$$
and
$$\sum_{k=0}^m2^{2k-1} = 2^{-1}\sum_{k=0}^m\left(2^2\right)^{k}$$
A: You ask for an approach to this kind of problem. The approach is: stop looking for a formula and think about techniques. The technique for dealing with $1+x^2+x^3+\dots+x^n$ is this. Suppose it is $S$. Now consider $xS$. That is the same except that it has lost a term at the front and gained one at the back. So it is now easy to write down a formula for $S$. 
How do you apply that to $x+x^3+x^5+\dots+x^{2n+1}$? What could you multiply the sum by to give something almost the same. Answer, multiply by $x^2$.
Of course, it is useful to know some formulae by heart. That saves time and it also gives you something to work from. But it is no good just saying I don't know a formula that exactly fits the new problem, so I will give up. The question is whether the new problem is similar to something I already know. If it is how could I adapt that ...
A: [EDIT] Thanks for accepting. This is an even more straightforward approach:
$$\begin{align}
S&=2^0+2^2+2^4+ \cdots +2^{2m}\\
&=4^0+4^1+4^2+\cdots +4^m\\
&=\frac 13 (4^{m+1}-1)\end{align}$$
$$\begin{align}
S'&=2^1+2^3+2^5+ \cdots +2^{2m+1}\\
&=2(4^0+4^1+4^2+\cdots +4^m)\\
&=\frac 23 (4^{m+1}-1)\end{align}$$

[ORIGINAL ANSWER] Consider the case where $n$ is even, i.e. $n=2m$. 
Let 
$$\begin{align}
S&=2^0+2^2+2^4+\cdots +2^{2m}\\
2S&=2^1+2^3+2^5+\cdots+2^{2m+1}\end{align}$$
Adding:
$$\begin{align}3S&=2^0+2^1+2^2+2^3+2^4+\cdots+2^{2m+1}\\
&=\sum_{r=0}^{2m+1}2^r\\
&=2^{2m+2}-1\\
S&=\frac 13 \left(2^{2m+2}-1\right)\\
&=\frac 13 \left(4^{m+1}-1\right)\end{align}$$

Similarly, for the case where $n$ is odd, i.e. $n=2m+1$, 
$$\begin{align}S'&=2^1+2^3+2^5+\cdots+2^{2m+1}\\
&=2(2^0+2^2+2^4+\cdots +2^{2m})\\
&=2S\\
&=\frac 23 \left(4^{m+1}-1\right)\end{align}$$
