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?

Let the terms of the sum be $a_n$. You need to find:
$$a_{n+1}+a_{n+2}+\cdots+a_{2n}=\\
(a_1+\cdots+a_{n}+a_{n+1}+\cdots+a_{2n})-(a_1+\cdots+a_n)=\\
S_{2n}-S_n$$
In your first method, you are subtracting the term $a_{n+1}$ and losing it.
Addendum: Note the correct formula to use is:
$$S_n=\sum_{k=1}^n k(k+1)=\frac{2n^3+6n^2+4n}{6}$$
Now consider the difference:
$$\sum_{r=n+1}^{2n}  r(r+1)=S_{2n}-S_n=\\
\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}=\\
\frac{7}{3}n^3+3n^2+\frac{2}{3}n.$$
