Find sum of noninteger number of terms How do you add a nonintegral number of terms together? For instance:
$$\sum_{i=0}^{2.5} 1+2i$$
All I can think of is:
$$(1+2*0) + (1+2*1) + (1+2*2) + (1+2*?)$$
I am a beginner in math, so be patient.
Edit:
I am trying to express $x^2$ in a different way, because the following works when $x$ is an integer:
$$x^2 = \sum_{i=0}^{x-1} 1+2i$$
Also, the expected output to this particular expression is $12.25$.
 A: The common interpretation of $$\sum_{i=0}^a f(i)$$ with $a\in\mathbb{R}$ is
\begin{align*}
\sum_{i=0}^a f(i)=f(0)+f(1)+f(2)+\cdots+f\left(\lfloor{a-1}\rfloor\right)+f\left(\lfloor{a}\rfloor\right)
\end{align*}
with $\lfloor a\rfloor$ the floor-function, denoting the greatest integer less or equal to $a$.

In the current case we have
  \begin{align*}
\color{blue}{\sum_{i=0}^{2.5}(1+2i)}&=(1+2\cdot0)+(1+2\cdot 1)+(1+2\cdot 2)\\
&=1+3+5\\
&\color{blue}{\,=9}
\end{align*}

A word of warning: The sigma-notation $\Sigma$ and the addition symbol $+$ share the same precedence level.

We have to use brackets if the term $2i$ should be summed up with the Sigma symbol,
  otherwise the following is valid
  \begin{align*}
\sum_{i=0}^{2.5}1+2i=\left(\sum_{i=0}^2 1\right)+2i=3+2i
\end{align*}
  which   is   usually not intended.

Hint: The scope of the sigma operator $\Sigma$ is solely defined via arithmetic precedence rules. The scope is given by the expression that follows immediately the $\Sigma$ and is valid respecting the arithmetic precedence rules up to an operator with precedence level equal to '$+$' or up to the end if no such operator follows.
You might find Chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik useful. It provides a thorough introduction in the usage of sums.
A: If you want to express $x^2$ in terms of a summation, you can do
$$\sum_{i=1}^{x} (2i-1) = \sum_{i=0}^{x-1} (1+2i) = x^2$$
(However, this has limited use, because in order to evaluate it at non-integer values of $x$, you need to expand the summation into $x^2$ and calculate it from there, which defeats the purpose of the summation for values such as $x=2.5$.)
