<s>[INST] Want help for prove formula by combinatoric argument $1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)=\frac{n(n+1)(n+2)}{3}$ 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?
 [/INST] 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:$$
 \frac{n(n+1)(n+2)}{3}= {(n+1)(n)} + {(n-1)(n)} + \ldots + {1\cdot 2}
$$
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