What is the closed form of $\sum^n_{k=0}(5^k+ k^3) $? What is the closed form of $
\displaystyle\sum^n_{k=0}(5^k+ k^3)
$?
We were given 6 basic summation rules and the closest one that we could work with doesn't fit the prperties. Thank you.
 A: Note that since you are looking at finite sum, there is no issue with convergence of series.
$$\begin{align*}\sum_{k=0}^n {5^k+k^3}&=\sum_{k=0}^n 5^k+\sum_{k=0}^nk^3\\&=\dfrac{5^{n+1}-1}{4}+\left(\dfrac{n(n+1)}{2}\right)^2\end{align*}$$
So, how did I get that?
The first summation on RHS is a geometric series with first term $5^0=1$ and common ratio $5$.
The second one is the sum of first $n$ cubes of natural numbers! 
For the first one,
\begin{equation*}
\sum_{k=m}^n {a^k}= 
\begin{cases} \dfrac{a^m(a^{n-m+1}-1)}{a-1} & \text{if $a>1$,}
\\
\\
(n-m+1)a &\text{if $a=1$,}
\\
\\
\dfrac{a^m(1-a^{n-m+1})}{1-a} & \text{if $a<1$,}
\end{cases}
\end{equation*}
So, I have used for the second sum, $$\sum_{k=1}^n {k^3}=\left(\dfrac{n(n+1)}{2}\right)^2$$
The other you'll find useful are (you might have been given the same formulae!)
$$\sum_{k=1}^n{k}=\dfrac{n(n+1)}{2}$$
$$\sum_{k=1}^n{k^2}=\dfrac{n(n+1)(2n+1)}{6}$$
And, the splitting is justified, only because the sum is finite! Otherwise, I cannot do this without checking for some criterion!
A: As suggested by Sp3000, split the sum as $\displaystyle\sum_{k=0}^{n} 5^k$ and $\displaystyle\sum_{k=0}^{n} k^3$. 
As you have said, the closed form for the sum of $k^3$ given to you and you can evaluate the second sum. 
Call the first sum $S=\displaystyle\sum_{k=0}^{n} 5^k=1+5+\cdots+5^n$. Then $5S=\displaystyle\sum_{k=1}^{n+1} 5^k=5+5^2+\cdots+5^{n+1}$. Therefore, 
$$5S-S=(5+5^2+\cdots+5^{n+1})-(1+5+\cdots+5^n)=5^{n+1}-1,$$
which implies that 
$S=(5^{n+1}-1)/4.$
