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.
 A: $\bf Hint:$ $\sum_{i=1}^n i(i+1)=\sum_{i=1}^n i^2+\sum_{i=1}^n i$. 
A: Hint: Note that the general term in your series is $2\binom{n+1}{2}$.
From the definition of Pascal's Triangle, we get
$$
\binom{n+2}{3}=\binom{n+1}{2}+\binom{n+1}{3}
$$
which leads to the formula, ripe for telescoping series:
$$
\binom{n+1}{2}=\binom{n+2}{3}-\binom{n+1}{3}
$$
A: I see two approaches:


*

*You can decompose it into (1²+2²+...+n²) + (1+2+...+n). For both of them formulas expressing the sum directly are easily available.

*Since your terms are quadratic, the sum can be expressed by a polynomial of third degree.
So you can use the ansatz a*x³ + b*x² + c*x + d and determine a, b, c, d so it fits 4 manually calculated elements.


You should be able to figure out the details from that. Both approaches work on other, similar problems too. So you should have them in your toolbox for later problems/the exam.
A: Another hint: $$k(k+1)=\frac{1}{3}(k(k+1)(k+2)-(k-1)k(k+1)).$$  The right-hand side gives you a telescoping sum.
A: If you don't know where to start, I suggest you rewrite the sum using the Σ(i:1-> n) notation. It eventually leads to a simpler expression.
A: 
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.
