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?
 A: 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:
$$\sum_{r=n+1}^{2n}  r(r+1)=S_{2n}-S_n=\\
\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}=\\
\frac{7}{3}n^3+3n^2+\frac{2}{3}n.$$
A: 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.
A: Another approach: it's clear that the result is a polynomial of $n$ of degree $3$, let $$\sum\limits_{r=n+1}^{2n}(r^2+r)=An^3+Bn^2+Cn+D=P(n)$$
thus
\begin{align*}P(n)-P(n-1)&=\sum\limits_{r=n+1}^{2n}(r^2+r)-\sum\limits_{r=n}^{2n-2}(r^2+r)\\
&=-(n^2+n)+((2n-1)^2+(2n-1))+((2n)^2+2n)\\
&=7n^2-n\\
&\equiv A(3n^2-3n+1)+B(2n-1)+C\\
&=3An^2+(-3A+2B)n+(A-B+C)
\end{align*}
$$
\begin{cases}
3A=7\\
-3A+2B=-1\\
A-B+C=0\\
A+B+C+D=P(1)=2^2+2=6
\end{cases}$$
$$
\begin{cases}
A=\frac{7}{3}\\
B=3\\
C=\frac{2}{3}\\
D=0
\end{cases}$$
Thus $$\sum\limits_{r=n+1}^{2n}(r^2+r)=\frac{7}{3}n^3+3n^2+\frac{2}{3}n.$$
