What is the closed form of ∑ k = 0 n ( 5 k + k 3 ) ?

Question

What is the closed form of             ∑              k        =        0            n        (          5      k        +          k      3        )   ?

hopeloothab9m · Accepted Answer

Note that since you are looking at finite sum, there is no issue with convergence of series.

\begin{aligned} \sum_{k = 0}^{n} 5^{k} + k^{3} & = \sum_{k = 0}^{n} 5^{k} + \sum_{k = 0}^{n} k^{3} \\ = \frac{5^{n + 1} - 1}{4} + {(\frac{n (n + 1)}{2})}^{2} \end{aligned}

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,

\sum_{k = m}^{n} a^{k} = {\begin{cases} \frac{a^{m} (a^{n - m + 1} - 1)}{a - 1} & if a > 1, \\ (n - m + 1) a & if a = 1, \\ \frac{a^{m} (1 - a^{n - m + 1})}{1 - a} & if a < 1, \end{cases}

So, I have used for the second sum,

\sum_{k = 1}^{n} k^{3} = {(\frac{n (n + 1)}{2})}^{2}

The other you'll find useful are (you might have been given the same formulae!)

\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}

\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 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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