Finding a sum as function of $n$ using summation formulas

Question

i need to find the sum as a function of $n$ using summation formulas of the following serie:

\begin{equation} \sum_{k=1}^n\bigg(4+\cfrac{2k}{n}\bigg)^2\bigg(\cfrac{1}{n}\bigg). \end{equation}

Any idea how to proceed? Any help will be appreciated!

Answer 1

I would proceed as follow:

$$\frac{1}{n}\sum_{k = 1}^n\ \frac{(4n^2+2k)^2}{n^2} = \frac{4}{n^3}\sum_{k = 1}^n (2n^2+k)^2 = \frac{4}{n^3}\sum_{k = 1}^n\ 4n^4 + k^2 + 4k$$

Now split the sum into three pieces, knowing that those are well known series:

$$\sum_{k = 1}^n 4n^4 = 4n^4\sum_{k = 1}^n = 4\cdot n^5$$

$$\sum_{k = 1}^n 4k = 4\sum_{k = 1}^n k = 4\frac{n(n+1)}{2} = 2n(n+1)$$

$$\sum_{k = 1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$

Combine the whole terms obtaining the final sum:

$$\frac{4}{n^3}\left(4n^5 + 2n(n+1) + \frac{n(n+1)(2n+1)}{6}\right)$$

Getting finally:

$$\frac{2 \left(38 n^2+15 n+1\right)}{3 n^2}$$

Answer 2

prove by induction that the given sum is equal to $$\frac{2 \left(38 n^2+15 n+1\right)}{3 n^2}$$