How to get this summation in closed form?

Question

$$\sum_{k = 0}^{n} \left( \dfrac{1}{n} - \dfrac{k^{2}}{n}\right)$$

How would one go about putting a summation in closed form if the lower limit is 0 instead of 1? I know you can't immediately use the summation formulas.

Answer 1

\begin{align} & \sum_{k = 0}^{n} \left( \dfrac{1}{n} - \dfrac{k^{2}}{n}\right) \\ = & \sum_{k = 0}^{n} \dfrac{1}{n} - \sum_{k = 0}^{n} \dfrac{k^{2}}{n} \\ = & \underbrace{\dfrac{1}{n} + \dfrac{1}{n} + \ldots + \dfrac{1}{n}}_{n+1 \text{ times}} - \dfrac{1}{n} \sum_{k = 0}^{n} k^{2} \\ = & 1 + \dfrac{1}{n}- \dfrac{1}{n} \times \dfrac{n (n + 1)(2n + 1)}{6}\\ = & 1 + \dfrac{1}{n} - \dfrac{(n + 1)(2n + 1)}{6} \\ = & \dfrac{-2n^{3} - 3n^{2} + 5n + 6}{6n} \end{align}

See this for the details about the squared sum of first $k$ natural numbers.

Here is a solution by Mathematica:

Answer 2

It can be factored: \begin{align} \sum_{k = 0}^{n} \biggl( \dfrac{1}{n} - \dfrac{k^{2}}{n}\biggr)&=\frac 1n \biggl(\sum_{k = 0}^{n}1-\sum_{k = 0}^{n}k^2\biggr)=\frac1n\biggl(n+1-\frac{n(n+1)(2n+1)}6\biggr)\\[1ex] &=\frac{(n+1)(-2n^2-n+6)}{6n}. \end{align} Now the polynomial $\;2n^2+n-6$ has an integer root: $-2$ and the product of the roots is $\dfrac{6}{-2}=-3$, so the other root is $\dfrac32$, and we have a factorisation: $$\sum_{k = 0}^{n} \biggl( \dfrac{1}{n} - \dfrac{k^{2}}{n}\biggr)= -\frac{(n+1)(n+2)(2n-3)}{6n}.$$