Summation of polynomial expression

Question

I'm looking for some assistance to the following problem:

$$\sum_{k=3}^{n}(k^2 - 3) = \sum_{k=3}^{n}{k^2} - \sum_{k=3}^{n}{3}$$ $$= \frac{n(n+1)(2n+1)}{6} - 3n$$

However, I know the last term is incorrect, as the answer is given as

$$\frac{n(n+1)(2n+1)}{6} - 3n + 1$$

Knowing the identity of $\sum_{i=1}^{n}3 = 3n$, how do I arrive at the last term?

Thanks.

Answer 1

Using the identity $$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6},$$ gives $$1+4+\sum_{i=3}^n i^2=\frac{n(n+1)(2n+1)}{6}.$$ and hence $$\sum_{i=3}^n i^2=\frac{n(n+1)(2n+1)}{6}-5.$$ We also have $$\sum_{i=3}^n 3=3n-6.$$ Substracting the one of these from the other gives the desired answer.

Answer 2

we have $\sum_{k=3}^n 3=3\sum_{k=3}^n 1=3(n-2)$

Answer 3

Note that $$\sum_{k=\color{blue}{3}}^{n}k^2\not=\frac{n(n+1)(2n+1)}{6},\ \ \ \sum_{k=\color{blue}{3}}^{n}3\not=3n$$ and that $$\sum_{k=\color{red}{1}}^{n}k^2=\frac{n(n+1)(2n+1)}{6},\ \ \ \sum_{k=\color{red}{1}}^{n}3=3n.$$ Hence, we have

$$\begin{align}\sum_{k=\color{blue}{3}}^{n}(k^2-3)&=\left(\sum_{k=\color{red}{1}}^{n}(k^2-3)\right)-\left(\sum_{k=1}^{2}(k^2-3)\right)\\&=\left(\frac{n(n+1)(2n+1)}{6}-3n\right)-(-2+1)\\&=\frac{n(n+1)(2n+1)}{6}-3n+1.\end{align}$$

Answer 4

your sum is given by $\sum_{k=3}^n k^3-3=\frac{1}{6}(n-2)(2n^2+7n-3)$