What is the closed form of $\sum^n_{k=0}(5^k+ k^3) $?

Question

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.

Answer 1

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!

Answer 2

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 $S=(5^{n+1}-1)/4.$