# Sum of powers for a non integer [duplicate]

I know that there are formulas for the first n squares,cubes,..., but is there a formula for the first n powers of a single non-integer?

For example, $$\sum_{i=0}^{n}1.1^i$$

Does a formulas to compute this sum exist or do i have to manually go in and find each power?

That is what is known as a geometric series. There is a closed formula for the finite sum

$$S= \displaystyle\sum_{k=0}^n a^k$$

$$aS=\displaystyle\sum_{k=1}^{n+1}a^k$$

If we subtract the first sum above from the second, we get

$$aS-S=a^{n+1}-1$$

$$S=\frac{a^{n+1}-1}{a-1}$$

This last division is only valid if $a\neq 1$.

So, in general, if you want to find a finite sum:

$$\displaystyle\sum_{k=0}^n a^k=\begin{cases}n \qquad\qquad\text{if}\qquad a=1\\ \frac{a^{n+1}-1}{a-1}\qquad \text{if}\qquad a\neq 1\end{cases}$$

In your example, $a=1.1$, so we have

$$\displaystyle\sum_{i=0}^n 1.1^i = \frac{1.1^{n+1}-1}{1.1-1} = 10\left(1.1^{n+1}\right)-10=\boxed{\frac{11^{n+1}}{10^n}-10}$$

If you want you could expand $1.1^{n+1}$ using the binomial theorem:

$$\displaystyle\sum_{i=0}^n 1.1^i = \displaystyle\sum_{k=0}^{n+1}\left(\frac{1}{10^{k-1}}\binom{n+1}{k}\right)$$

where $\binom{n+1}{k}$ is a binomial coefficient.