An expression of $1\cdot2 + 2\cdot3 + \cdots + n\cdot(n+1)$ I got a question in my homework, which is:

Find the following sum and prove your claim:
      $$1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + n\cdot(n+1).$$

I want to prove this by mathematical induction, but I couldn't find an expression of the sum. If anyone has any idea, please share with me. Thank you.


Find the following sum and prove your claim:
$$1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + n\cdot(n+1).$$

$$1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + i(i+1) + \cdots + n(n+1) = \sum_{i=1}^n i(i+1)=\sum_{i=1}^n i^2+\sum_{i=1}^n i \tag1$$
It is easy to prove
$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
But how you derive $\sum_{i=1}^n i^2$ here? Expand $(i-1)^3$:
$$(i-1)^3 = i^3 - 3i^2 + 3i -1 \ \Rightarrow \ i^3 -(i-1)^3 = + 3i^2 - 3i +1 \tag 2$$
Thus,
$$\sum_{i=1}^n \left(i^3 -(i-1)^3\right) = \sum_{i=1}^n  3i^2 - \sum_{i=1}^n 3i + \sum_{i=1}^n 1 \tag3$$
The RHS of the equation $(3)$ is telescopic series, sum of which is eqial to $n^3$. Thus,
$$3\sum_{i=1}^n  i^2 = n^3 + 3\sum_{i=1}^n i - \sum_{i=1}^n 1 = n^3 +3 \frac{n(n+1)}{2} -n$$
$$\therefore \ \sum_{i=1}^n  i^2 = \frac{1}{3} \left(n^3 +3 \frac{n(n+1)}{2} -n \right) = \frac{n}{6} \left(2n^2 +3 n + 1 \right) \\= \frac{n(2n+1)(n+1)}{6}$$
$$\therefore \ \sum_{i=1}^n i(i+1)=\sum_{i=1}^n i^2+\sum_{i=1}^n i = \frac{n(2n+1)(n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3} $$
To prove this is correct, check for $n=1$: $\sum_{i=1}^n i(i+1)= 1 \times 2 = 2$ and $\frac{n(n+1)(n+2)}{3} = \frac{1(1+1)(1+2)}{3} =2$, thus answer is true for $n=1$.
Therefore, soppose it is true for $n=k$. If it is true for $n=k+1$, the difference in summation should be $(k+1)(k+2)$:
$$\sum_{i=1}^{k} i(i+1)= \frac{k(k+1)(k+2)}{3} \tag4$$
$$\sum_{i=1}^{k+1} i(i+1)= \frac{(k+1)(k+2)(k+3)}{3} \tag5$$
The equations $(6)-(5)$:
$$\sum_{i=1}^{k+1} i(i+1) - \sum_{i=1}^{k} i(i+1) = \frac{(k+1)(k+2)(k+3)}{3} - \frac{k(k+1)(k+2)}{3}\\ = \frac{(k+1)(k+2)(k+3 -k)}{3} = (k+1)(k+2)$$
Thus solution is proved to be correct.
