<s>[INST] Evaluating Sum at bounds I have to find an expression in terms of n using standard results for $$\sum_{r=n+1}^{2n}  r(r+1)$$
And have found the general equation
$$\sum_{r=n+1}^{2n}  r(r+1) = \frac{2n^3+6n^2+4n}{6}$$
However evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n+1)^3+6(n+1)^2+4(n+1)}{6}$$
does not yield the correct answer, yet evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}$$
gives the correct answer
Im at a loss here, why am I not getting the correct answer by finding the difference of the sum between the two bounds?
 [/INST] Your general equation should be
$$\begin {align} \sum_{r=n+1}^{2n}  r(r+1)&=\sum_{r=n+1}^{2n}  (r^2+r)\\
&=\sum_{r=1}^{2n}  (r^2+r)-\sum_{r=1}^{n}  (r^2+r)\\
&=\frac 16\left((2n)(2n+1)(4n+1)\right)+\frac 12\left(2n(2n+1)\right)-\frac 16\left((n)(n+1)(2n+1)\right)+\frac 12\left(n(n+1)\right)\\
&=\frac 16\left(16n^3+12n^2+2n\right)+\frac 12\left(4n^2+2n)\right)-\frac 16\left(2n^3+3n^2+n\right)+\frac 12\left(n^2+n\right)\\
&=\frac 16\left(14n^3+9n^2+n\right)+\frac 12\left(3n^2+n)\right)\\
&=\frac 16\left(14n^3+18n^2+4n\right)
\end {align}$$
which does not match your result and checks with Alpha.
</s>