Want help for prove formula by combinatoric argument $1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)=\frac{n(n+1)(n+2)}{3}$

Question

I have a problem.I just read first combinatorics textbook and I found the identity $1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)=\frac{n(n+1)(n+2)}{3}$. I can prove this by induction,but the problem is from combinatorics textbook and I have no idea how to begin the proof. How can I prove it by combinatorics argument?

Answer 1

On the right side, you have the number of ways of selecting three items without replacement from a set of $n+2$ distinct items, multiplied by $2$. For simplicity, I'll assume there are $n$ items first, then I'll up the index later.

Now, suppose you certainly select the first item. The number of ways of choosing the rest are $\frac{(n-1)(n-2)}{2}$. Otherwise, you don't select the first item but rather the second one, and here the number of possibilities are $\frac{(n-3)(n-2)}{2}$. Repeating, you can finally select the third item from last, which gives $\frac{1\cdot 2}{2}$.These exhaust all the cases, so you have the equation: $$ \binom{n}{3} = \frac{(n-1)(n-2)}{2} + \frac{(n-3)(n-2)}{2} + \ldots + \frac{1\cdot 2}{2}. $$

Multiply by two and you have your formula: $$ \frac{n(n-1)(n-2)}{3}= {(n-1)(n-2)} + {(n-3)(n-2)} + \ldots + {1\cdot 2} $$

Now, just put $n+2$ in place of $n$ for:$$ \frac{n(n+1)(n+2)}{3}= {(n+1)(n)} + {(n-1)(n)} + \ldots + {1\cdot 2} $$

Answer 2

Note that RHS is $$2\binom{n+2}3$$

Cancel this $2$ and put it as a denominator on each term of the LHS and you get

$$\sum_{k=2}^{n+1}\binom k2=\binom{n+2}3$$

RHS: How many subset of $3$ elements has the set $\{1,\ldots,n+2\}$?

LHS: For each $k$, how many subsets of $3$ elemets whose maximum is $k+1$ has the set $\{1,\ldots,k+1\}$?