# 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: $1^2+3^2+5^2 = 35$ ?

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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.

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

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