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$ \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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  • $\begingroup$ Sorry, I mean notation for general situation. If there will be not $ i^2 $ but $ f(i) $ $\endgroup$ Aug 29 '12 at 15:34
  • $\begingroup$ Second variant suits me if this notation is acceptable, thank you! :) $\endgroup$ Aug 29 '12 at 15:34
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    $\begingroup$ Usually when conditions are added the upper index is not used, so the subscript would be $1\le i\le 5,\ i\text{ odd}$. $\endgroup$
    – Charles
    Aug 29 '12 at 15:47
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    $\begingroup$ @Charles: I agree. Your way is better. $\endgroup$
    – Thomas
    Aug 29 '12 at 15:55
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    $\begingroup$ @Thomas you have typo in your current answer: $ 1 \leq 1\leq 5 $ should probably be $ 1 \leq i\leq 5 $ :) $\endgroup$ Aug 29 '12 at 16:06
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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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  • 6
    $\begingroup$ With the little caveat that $2i$ is not odd. $\endgroup$
    – Fabian
    Aug 29 '12 at 15:47

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