How to sum $\sum_{k=1}^n (k+1)(k)(k-1)$ Is there an intelligent way to do this sum without using sums of cubes and sums of squares?
$$\sum_{k=1}^n (k+1)(k)(k-1)$$
 A: There is a discrete version of calculus, known as the difference calculus, which studies infinite sequences $f:\mathbb Z^+ \to \mathbb R$ (instead of studying functions from $\mathbb R$ to $\mathbb R$, as in calculus).
The "first difference" of a sequence $f:\mathbb Z^+ \to \mathbb R$ is denoted $Df$ and is the infinite sequence defined by:
\begin{equation*}
(Df)_k = f_{k+1} - f_k.
\end{equation*}
The first difference is a discrete version of the derivative.
The equation
\begin{equation*}
\sum_{k=1}^N (Df)_k = f_{N+1} - f_1
\end{equation*}
is a discrete version of the fundamental theorem of calculus, and is easily proved by noting that the sum on the left telescopes.
With this theorem, finding the sum of the first $N$ terms of a sequence is reduced to the problem of finding an "anti-difference" of the sequence
(just as evaluating integrals is reduced to finding antiderivatives in calculus).
Given a positive integer $k$, define $k^{\underline{m}}$ by
\begin{equation*}
k^{\underline{m}} = k(k-1)(k-2)\cdots(k-m + 1).
\end{equation*}
For example,
\begin{equation*}
k^{\underline{3}} = k(k-1)(k-2).
\end{equation*}
You can easily prove that if $f_k = k^{\underline{m}}$, then
\begin{equation*}
(Df)_k = m k^{\underline{m-1}}.
\end{equation*}
This is a discrete version of the power rule from calculus.
Now suppose $g_k = k^{\underline{3}}$.
From the discrete power rule, an anti-difference of $g$ is the sequence
\begin{equation*}
f_k = \frac{k^{\underline{4}}}{4}.
\end{equation*}
This is just like finding an antiderivative of $x^3$ in calculus.
This allows us to use the discrete fundamental theorem:
\begin{align*}
\sum_{k=1}^N k(k-1)(k-2) &= \sum_{k=1}^N k^{\underline{3}} \\
&= \sum_{k=1}^N (Df)_k \\
&= f_{N+1} - f_1 \\
&= f_{N+1} \\
&= \frac{(N+1)N(N-1)(N-2)}{4}.
\end{align*}
A: Notice 
$$\require{cancel}(k-1)k(k+1) = \frac{(k-1)k(k+1)\color{red}{(k+2)} - \color{blue}{(k-2)}(k-1)k(k+1)}{\color{red}{(k+2)}-\color{blue}{(k-2)}}$$
You can rewrite your sum as a telescoping one...
$$\begin{align}
\sum_{k=1}^N (k-1)k(k+1) 
&= \frac14 \left((N-1)N(N+1)(N+2) - (1-2)(1-1)1(1+1)\right)\\
&= \frac14 (N-1)N(N+1)(N+2)
\end{align}$$
A: HINT:
Let $(k-1)k(k+1)=T(k+1)-T(k)$ where $T(k)=a_0+a_1k+a_2k^2+a_3k^3+a_4k^4+\cdots$
$\implies (k-1)k(k+1)=k^3-k=a_1[(k+1)-1]+a_2[(k+1)^2-k^2]+a_3[(k+1)^3-k^3]+a_4[(k+1)^4-k^4]+\cdots$
Clearly  Comparing the coefficients of$x^n, n\ge5;a_n=0\forall n\ge5$
Comparing the coefficients of $k^3,4a_4=1$
Comparing the coefficients of $k^2,0=6a_4+3a_3\iff a_3=\cdots$
and so on
Now see Telescoping Series
A: The identity
$$
\sum_{n=k}^m\binom{n}{k}=\binom{m+1}{k+1}
$$
in the case $k=3$ says that
$$
\sum_{k=2}^n6\binom{k+1}{3}=6\binom{n+2}{4}
$$
Therefore,
$$
\sum_{k=2}^n(k+1)k(k-1)=\frac{(n+2)(n+1)n(n-1)}4
$$

The identity mentioned above is proven in this answer, using $a=k$, $b=0$, and $n=m$.
A: Generally, a problem of this sort can be solved by using the identity $$(q-p+2)\prod_{j=p}^q(k+j)=\prod_{j=p}^{q+1}(k+j)-\prod_{j=p-1}^q(k+j).$$This is easily proved by taking out the common factor $\prod_{j=p}^q(k+j)$ from the products on the right-hand side and simplifying. For example, if $p=-1$ and $q=1,$ this means that $$4(k-1)k(k+1)=(k-1)k(k+1)(k+2)-(k-2)(k-1)k(k+1).$$When we sum expressions like this from $k=1$ to $n$, say by writing them in a column and adding, notice that each subtracted term on the right cancels the initial term on the line above it, so that only the initial term in the bottom row and the subtracted term in the first row remain:$$4\sum_{k=1}^n(k-1)k(k+1)=(n-1)n(n+1)(n+2)-(1-2)(1-1)1(1+1)$$or $\sum_{k=1}^n(k-1)k(k+1)=\frac14(n-1)n(n+1)(n+2).$ 
