Determine Sum of $\sum_{k=3}^{25}{k^2-5k+9}$ Given $\sum_{k=3}^{28}(k-3)^2 = 14,910$ and $\sum_{k=0}^{25}k = 325$ I am working through some questions from a textbook which states that I should determine the sum without expanding or calculating any sums. I have been given the following information.
$\sum_{k=3}^{28}(k-3)^2 = 14,910$ and $\sum_{k=0}^{25}k = 325$
And must calculate the sum for $\sum_{k=3}^{25}{k^2-5k+9}$
So far I've noticed the following,
${\sum}_{k=3}^{28}\left(k-3\right)^{2}={\sum}_{k=0}^{25}k^{2}$
${\sum}_{k=0}^{25}k={\sum}_{k=3}^{28}k-3$
I'm also noticed that, $\left(k-3\right)^{2}=k^{2}-6k+9$, is very close to the sum I need to find, but am not sure how this helps.
I calculated the sum as follows, but I think the textbook would not have wanted me to calculate it in this way,
$\sum_{k=3}^{25}{k^2-5k+9}={\sum}_{k=3}^{25}k^{2}-5{\sum}_{k=3}^{25}k+{\sum}_{k=3}^{25}9=14,910-5-5\times\left(325-3\right)+22\times9=13,493$
I believe my answer is wrong however I'm not sure how to proceed. I would appreciate it if someone could provide some sort of hint to guide me in the right direction.

We can manually subtract terms from each of the two sums to get the desired range:
$$\sum_{k=3}^{25}(k-3)^2=14910-25^2-24^2-23^2=13180$$
$$\sum_{k=3}^{25}k=325-0-1-2=322$$
The final answer is $13180+322=13502$.

In fact, the given values are completely wrong; the actual correct answer is $4117$.
