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