How do I evaluate $\sum_{r=1}^{n} [r(r+1)(r+2)(r+3)] $? I need to evaluate:
\begin{equation}
\sum_{r=1}^{n} [r(r+1)(r+2)(r+3)]
\end{equation}
But I don't know where to start. Can someone possibly give a hint?
I need to solve this using telescopic series.
 A: $r(r+1)(r+2)(r+3) = 24\binom{r+3}{4}$, and since
$$ \sum_{k=j}^{N}\binom{k}{j}=\binom{N+1}{j+1}$$
holds by induction, we have:
$$ \sum_{r=1}^{n} r(r+1)(r+2)(r+3) = 24\sum_{r=1}^{n}\binom{r+3}{4} = 24\binom{n+4}{5} = \frac{n(n+1)(n+2)(n+3)(n+4)}{5}.$$
A: $$
r(r+1)(r+2)(r+3) = \frac15\left((r+1)(r+2)(r+3)(r+4)(r+5)- r(r+1)(r+2)(r+3)(r+4)\right) \\
$$
A: I suggest expanding the product to get $$\sum_{r=1}^n(r^4+6r^3+11r^2+6r) = \\ =\sum_{r=1}^n r^4 + 6\sum_{r=1}^n r^3 + 11\sum_{r=1}^n r^2 + 6\sum_{r=1}^n r$$
Now the problem is finding formulas for these sums. The linear sum you probably know (or should) and you might know the sum of squares too. For the ones you don't, it's useful to know that the sum $\sum_{r=1}^n r^k$ is a polynomial in $n$ of degree $k+1$. Therefore if you calculate the first $k+1$ sums, you'll get a linear system of equations in the coefficients (to be determined) of the polynomial.
For example, take $S(n) = \sum_{r=1}^nr^2$. We conjecture that this is equal to $an^3+bn^2 + cn + d$. First of all note that $d = 0$, necessarily. Next: $$\begin{align}S(1) &= 1 &= & a + b + c \\ S(2) &= 5 &= & 8a + 4b + 2c \\ S(3) &= 14 &= &27a + 9b + 3c\end{align}$$
Use your preferred method to solve the system. The same will go for the other sums.
This is quite a brute force method and as such it'll be tedious, but it's good to have in your arsenal of strategies. Additionally, you can skip the intermediate step of calculating the first $n$ squares, cubes, etc. by conjecturing from the very beginning that the sum is a polynomial of degree 5.
Edit: I now see you need to use the telescoping sum, so disregard if you wish. I will leave the answer anyway for any who might find it useful.
