<s>[INST] Find $S=2016^2 + 2015^2 +2014^2 -2013^2 -2012^2 -2011^2 .....+6^2 +5^2 +4^2-3^2-2^2-1^2$ Simplifying the expression as
$$2016^2+2015^2+2014^2+2013^2+2012^2+2011^2...6^2+5^2+4^2+3^2+2^2+1^2-2(2013^2+2012^2+2011^2 +2007^2+2006^2+ 2005^2....3^2+2^2+1^2)$$
Now the left part can be evaluated by using $\sum = \frac{(n)(n+1)(2n+1)}{6}$, but I am not sure about the right part. How should I solve that one?
 [/INST] Note that $2016=6\cdot 336.$ Then rewrite the whole sum as:
$$S = \sum_{k=1}^{336}\left((6k)^2+(6k-1)^2+(6k-2)^2-(6k-3)^2-(6k-4)^2-(6k-5)^2\right) = $$
$$ = \sum_{k=1}^{336}(108k-45)=...$$
Or, if you insist on your own method, then the thing in the bracket you are struggling with is simply:
$$T = \sum_{k=1}^{335}\left((6k+3)^2 + (6k+2)^2+(6k+1)^2\right)$$
</s>