How to evaluate the following summation I am trying to find the definite integral of $a^x$ between $b$ and $c$ as the limit of a Riemann sum (where $a > 0$):
$I = \displaystyle\int_b^c \! a^{x} \, \mathrm{d}x.$
However, I'm currently stuck in the following part, in order to find S:
$S = \displaystyle\sum\limits_{i=1}^n \displaystyle{a^{\displaystyle\frac{i(c-b)}{n}}}$
Is there a formula for this kind of expression? Thank you for your help.
 A: Yes, there is a formula. notice that $$a^{i(c-b)/n}=(a^{(c-b)/n})^i.$$ So, your sum is just a sum of a geometric progression.
A: Note that: 
$$\begin{align}
S &= \sum\limits_{i=1}^n \displaystyle{a^{\displaystyle\frac{i(c-b)}{n}}} \\
&= \sum\limits_{i=1}^n \left(a^{\left(\dfrac{(c-b)}{n}\right)}\right)^i
\end{align}$$
Now, we can use the finite form of the geometric series formula:
$$S = \frac{\left(a^{\left(\dfrac{(c-b)}{n}\right)}\right)-\left(a^{\left(\dfrac{(c-b)}{n}\right)}\right)^{n+1}}{1-a^{\left(\dfrac{(c-b)}{n}\right)}}$$
...and that limit will be pretty nasty, but do-able.  (I think.)
A: This might help.  Suppose $y = a^x$. Then $\ln(y) = x\ln(a)$. Differentiating you get
$${y'\over y} = \ln(a),$$
so $$ y' = y\ln(a) = a^x \ln(a).$$
Hence 
$$\int a^x\, dx = {a^x\over \ln(a)} + C.$$
A: Correct me if I'm wrong, but it looks like $a,b,c$ and $n$ are all constants inside that sum.  So
$$
S = \sum_{i=1}^n a^{\frac{i(c - b)}{n}} = \sum_{i=1}^n \left(a^{\frac{c - b}{n}}\right)^i
$$
which is just a finite geometric sum.  Letting $r = a^{\frac{c - b}{n}}$, this sums to $\frac{1 - r^{n+1}}{1 - r} - 1$.
