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For example: $\ x^2$ is the sum of the terms of $2x-1 $ (1+3+5+7)

Would that be written like: $\ \sum^{t}_{x=1} 2x-1=x^2$

$\ t$ being the whatever term.

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$\sum_{x=1}^t (2x-1)=t^2$. The parentheses are necessary, for $\sum_{x=1}^t 2x-1$ usually means add up the $2x$, subtract $1$ at the end. And the answer is $t^2$. – André Nicolas Sep 19 '12 at 23:34
Note that the free variable on the left is $t$, not $x$, so the result should be in terms of $t$ as André Nicolas has written it. – Ross Millikan Sep 20 '12 at 0:24
up vote 2 down vote accepted

$$\sum_{n=0}^{x-1} (2n+1)=1+3+5+\cdots+(2x+1)=x^2$$

because $n=0$ initially and the sum ends when $n=x$.

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Or you could write it as $$\sum_{n=1}^x (2n-1) = x^2$$ – Robert Israel Sep 20 '12 at 0:26

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