Is there any way to expand this using the binomial theorem? $(h-2.5i)^{1/2}$
I'm trying to isolate i, is it possible?
Cheers!
EDIT: $i$ is NOT $\sqrt{-1}$, it's just a variable and $h$ is a constant.
EDIT2: It's in sigma notation like so:
$$\sum_{i=1}^{14}(h-2.5i)^{1/2}$$
 A: Here is the most constructive answer I can give:
The expression
$\sum_{i=1}^{y}(h-ci)^{n}$ may have a closed form. Expansion would show:
$$\sum_{i=1}^{y}(h-ci)^{n}=(h-c)^n+(h-2c)^n+\dots+(h-(y-1)c)^n+(h-yc)^n$$
There is a problem here: The basic binomial theorem only applies to the case $(a+b)^n$ where $n \in \mathbb{Z}^{+}$. (I do not know the conditions on $a$ and $b$.)  Two different things are astray here: There is a coefficient greater than $1$ in every binomial but the first and in our specific case, $n$ is not a nonnegative integer.
These two issues are what makes this so hard to figure out. I can't seem to find any binomial series that works for the form $(h-ac)^{n}$ where $a \geq 1$ and $0<n<1$. Even Newton's generalized binomial series does not suffice, which really shows just how cumbersome this particular scenario is.
Looking for a non-general way of helping you is even more cumbersome upon inspection:
$$\sum_{i=1}^{14}(h-2.5i)^{\frac{1}{2}}=\sqrt{(h-2.5)}+\sqrt{(h-5)}+\dots$$
Do you see how this does not admit any sort of collecting of the terms in a more convenient fashion? You simply cannot add these terms and find a closed form. The only possible (slight) simplification of this involves turning the decimal into a fraction:
$$\sum_{i=1}^{14}\sqrt{h-\frac{5}{2}i}=\sum_{i=1}^{14}\sqrt{\frac{2h-5i}{2}}=\sum_{i=1}^{14}\frac{\sqrt{2h-5i}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\sum_{i=1}^{14}\sqrt{2h-5i}$$
You could then make life a little easier by rationalizing the denominator:
$$\frac{1}{\sqrt{2}}\sum_{i=1}^{14}\sqrt{2h-5i}=\frac{1}{2}\sum_{i=1}^{14}\sqrt{4h-10i}$$
The above avoids as many radicals as possible and converts the problem into an integer form.
In conclusion: There isn't a lot you can do to make your life easier. Sorry. :/
A: 
I'm trying to isolate i

Let's say you have:
$y=(h-2.5i)^{1/2}$
then
$y^2=h-2.5i$, hence:
$i= (h-y^2)/2.5$
Edit: 
You could use binomial theorem formula below but notice the condition associated. (source: PDF-Bionomial Theorem:

Edit 2:
A more generalized form is:

A: Whenever $|h / (2.5 i)| \le 1|$ you can use the binomial theorem, where:
$$
{1/2 \choose k} = \frac{(-1)^{k - 1}}{2^{2k - 1} \cdot k}
        \cdot \binom{2k - 2}{k - 1}
$$
Don't know if this helps much.
