Summation of $2^n$ where $n$ is even or odd

Question

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.

Are you sure that the first sum is to start with $2^0$ and not $2^1$? And didn't you mix up "even" and "odd"? — Frunobulax, Sep 08 '14 at 14:48
@Frunobulax Note that you can always compute the sum starting with $2^1$ and then add $2^0$ to the final result. — Surb, Sep 08 '14 at 14:52
@Surb: I know that. My question was if the OP knew what he was asking for. — Frunobulax, Sep 08 '14 at 15:01

score 4 · Answer 1 · answered Sep 08 '14 at 14:44

4

Hint: $$\sum_{k=0}^m2^{2k} = \sum_{k=0}^m\left(2^2\right)^{k}$$ and $$\sum_{k=0}^m2^{2k-1} = 2^{-1}\sum_{k=0}^m\left(2^2\right)^{k}$$

answered Sep 08 '14 at 14:44

Surb

55,662

1

Very neat answer. – Hypergeometricx Sep 08 '14 at 14:54
@hypergeometric Thanks :) – Surb Sep 08 '14 at 14:55
Well the index in the second is $k=1$ – kingW3 Sep 08 '14 at 15:31
@kingW3 Well my goal was to highlight the formula $\sum_{k=0}^m x^k = \frac{1-x^{m+1}}{1-x},x \neq 1$ (which seems to be know by the OP) in both cases. I think it is now the job of the OP to add/remove "extremal terms" in order to recover the sums he wants to compute. – Surb Sep 08 '14 at 15:44

score 2 · Answer 2 · answered Sep 08 '14 at 14:55

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 ...

Hypergeometricx · Accepted Answer · 2014-09-09T03:52:34.700

[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. $n=2m+1$,

$$\begin{align}S'&=2^1+2^3+2^5+\cdots+2^{2m+1}\\ &=2(2^0+2^2+2^4+\cdots +2^{2m})\\ &=2S\\ &=\frac 23 \left(4^{m+1}-1\right)\end{align}$$

But this summation cannot be a fraction right ? So, will it (the part except 1/3) be always divisible by 3 ? — divyenduz, Sep 08 '14 at 19:49
Yes, because $(4^{m+1}-1)=(4-1)(4^m+4^{m-1}+\cdots +1)=3(4^m+4^{m-1}+\cdots +1)$ is divisible by 3. — Hypergeometricx, Sep 09 '14 at 03:46

Summation of $2^n$ where $n$ is even or odd

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