# Sigma notation only for odd iterations

$\sum_{i=0}^{5}{i^2} = 0^2+1^2+2^2+3^2+4^2+5^2 = 55$

How to write this Sigma notation only for odd numbers: $1^2+3^2+5^2 = 35$ ?

You could write $$\sum_{i=1}^{3} f(2i-1).$$ Otherwise it is allowed to write $$\sum_{1 \leq i\leq 5, i \text{ odd}} f(i).$$ (Here in your example $f(i) = i^2$ of course).

So in general whatever condition you have on the index, you can write that underneath the sum. In general you will find some people prefer one thing over another.

• Sorry, I mean notation for general situation. If there will be not $i^2$ but $f(i)$ Aug 29 '12 at 15:34
• Second variant suits me if this notation is acceptable, thank you! :) Aug 29 '12 at 15:34
• Usually when conditions are added the upper index is not used, so the subscript would be $1\le i\le 5,\ i\text{ odd}$. Aug 29 '12 at 15:47
• @Charles: I agree. Your way is better. Aug 29 '12 at 15:55
• @Thomas you have typo in your current answer: $1 \leq 1\leq 5$ should probably be $1 \leq i\leq 5$ :) Aug 29 '12 at 16:06

Just use the following for any $f(i)$: $$\sum_{i=0}^n f(2i+1)$$

Edit: Sorry, I somehow mistook the question for "even".

• With the little caveat that $2i$ is not odd. Aug 29 '12 at 15:47