Sum $\sum_{k=1}^n{k(k-1)}$

Question

I have a series as

$$n(n-1)+(n-1)(n-2)+(n-2)(n-3)+ \ldots +n\ \text{terms}$$

How to provide a summation of this series? I think it would be similar to:

$$\sum_{i=0}^{N}(n-i)(n-(i+1)).$$

Answer 1

This is $$\sum_{k=1}^n{k(k-1)}=\sum_{k=1}^n{(k^2-k)}=\sum_{k=1}^n k^2-\sum_{k=1}^nk = \frac{n(n+1)(2n+1)}6-\frac{n(n+1)}2$$

Answer 2

Here is a combinatorial argument as lulu mentioned:

For each $n\geq 1$, define the set $S_n=\{1,2,3,\dots,n\}$. For a fixed $n\geq 1$, let us count the number of ways to choose a subset $S_k\subseteq S_n$, with $k\leq n$, and then designate two special elements $a$ and $b\in S_k$ as "best" and "second best".

We can count this quantity in two ways:

1. For each $k\leq n$, we have the set $S_k=\{1,2,3,\dots,k\}$. The number of ways to select the "best" element is $k$, and once that is chosen, the number of ways to select the "second best" element is $k-1$. Hence, there are $k(k-1)$ ways. Summing over each $k$ we have $\sum_{k=1}^{n}k(k-1)$.

2. Choose three distinct elements $a,b,c\in S_{n+1}$, and suppose without loss of generality $a<b<c$. There are ${n+1\choose 3}$ ways to do this. Let $k=c-1$, so that $k\leq n$. Then $a,b\in S_k$, so we may designate $a$ and $b$ to be "best" and "second best". There are two choices, either $a$ is "best" or $b$ is "best". Thus the number of ways is $2\times{n+1\choose 3}=\frac{n(n+1)(n-1)}{3}$.

Since we have counted the same quantity in two different ways, the results must be equal. Therefore, $\sum_{k=1}^{n}k(k-1)=2{n+1\choose 3}=\frac{n(n+1)(n-1)}{3}$.

Answer 3

One can see this as follows: Since $k(k-1) = \dfrac{(k+1)k(k-1) -k(k-1)(k-2)}{3} = a_{k+1}-a_k$ where $a_k = \frac{k(k-1)(k-2)}{3}$, one can see that \begin{align*} \sum_{k = 1}^n k(k-1) & = \sum_{k=1}^n ( a_{k+1}-a_k) \\ & = (a_{n+1} - a_n) + (a_{n} + a_{n-1}) + \cdots + (a_2 + a_1) \\ & = a_{n+1} - a_1 = (n+1)\cdot n \cdot (n-1)/3 \end{align*}

This method can induce the formulae for $f(a, n) = \sum_{k=1}^n k(k+1)\cdots(k+a-1)$. ($a \in \mathbb{N})$.

My flavor is first have $f(a, n)$ and use this to know what is $\sum_{k=1}^n k^a$, not the other direction.