I will tell a story. We have $n+2$ people, and their ages are $1$, $2$, $3$, $4$, and so on up to $n+2$.
We want to choose $3$ people from these $n+2$ people.
The number of ways to do this is
$$\binom{n+2}{3}=\frac{(n+2)(n+1)(n)}{3!}$$
Let us count the number of ways of choosing the $3$ people in another way.
Let us first count the number of ways to choose the $3$ people, if the youngest person chosen is to be the $1$ year-old. Then we must choose $2$ people to go with the $1$ year-old, and this can be done in $\binom{n+1}{2}$ ways.
Now count the number of ways to choose the people, if the youngest person chosen is to be $2$ years old. We must choose $2$ people from the $n$ people who are older than $2$. This can be done in $\binom{n}{2}$ ways.
Now count the number of ways to choose the people, if the youngest person chosen is to be $3$ years old. We must choose $2$ people from the $n-1$ people who are older than $3$. This can be done in $\binom{n-1}{2}$ ways.
Go on, and on. Finally, count the number of ways to choose $3$ people, if the youngest person chosen is $n$ years old. There is only $1$ way, of course, but for consistency I will call the number of ways $\binom{2}{2}$.
We conclude that
$$\binom{n+2}{3}=\binom{n+1}{2}+\binom{n}{2}+\binom{n-1}{2}+\cdots +\binom{2}{2}$$
Now $\binom{n+1}{2}=\frac{(n+1)(n)}{2!}$, and $\binom{n}{2}=\frac{(n)(n-1)}{2!}$, and so on.
Thus
$$\frac{(n+1)(n)}{2!}+ \frac{(n)(n-1)}{2!}+ \cdots +\frac{(2)(1)}{2!}=\frac{(n+2)(n+1)(n)}{3!}$$
Note that $2!=2$ and $3!=6$. Multiply both sides of our expression by $2$, and reverse the order of summation.
We get
$$\sum_0^n(i+1)(i)=\sum_1^n (i+1)(i)=\frac{(n+2)(n+1)(n)}{3}$$
So we have found an expression for our sum.
Comment: Almost exactly the same idea works in general, for computing related sums, such as $\sum_1^n (i+2)(i+1)(i)$. It turns out that finding sums of this type by the technique just described is far easier than, for example, multiplying $(i+2)(i+1)(i)$ out and using formulas for sums of powers. Combinatorial ideas are powerful!
It turns out that the easiest way to find formulas for sums of powers is to express these in terms of the sums we have just shown how to evaluate.
In some sense, a sum like $\sum_1^n (i+1)(i)$ is a more "natural" object than the more familiar $\sum_1^n i^2$.