# 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.

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What were you given and what have you tried? Also, it might be easier to split the sum up into two parts: $\sum_{k=0}^{n} 5^k$ and $\sum_{k=0}^{n} k^3$. Note that the first series is geometric. –  Sp3000 Jan 24 '12 at 6:56
We had the closed form for k^3 given to us, but the closest thing for the 5^k is either (ar^n+1=a)/r-1 , where a and r are real numbers and r !=1, or another closed form that doesn't work, because it requires 5 to be less than 1. I'm not entirely sure if it's fine to have it where 5 = a and r = 1 for the first example. –  Alex Crenell Jan 24 '12 at 7:03
Do you mean you were given something like: $$S_n = \frac{a(r^{n+1} - 1)}{r - 1}, r \not= 1$$ where $r$ is the ratio between successive terms in a geometric series and $a$ is the initial term? (Noting that there are $n+1$ terms between 0 and n inclusive) If so then a substitution of $r = 5$ and $a = 5^0 = 1$ can be done. –  Sp3000 Jan 24 '12 at 7:08

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!

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Thank you very much! I wasn't sure how to approach it. –  Alex Crenell Jan 24 '12 at 7:08

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

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