zmudz one year ago find polynomial f(n) such that for all integer n>= 1 we have 3(1 * 2 + 2* 3 + ... + n(n+1)) = f(n).

1. amoodarya

do you mean f(n)=$3(1*2+2*3+3*4+...+n(n+1))$?

2. amoodarya

$3(1*2+2*3+3*4+...+n(n+1))=3\sum_{1}^{n}i(i+1)=\\3\sum_{1}^{n}(i^2+i)=\\3\sum_{1}^{n}(i^2)+3\sum_{1}^{n}(i)=\\3(\frac{n(n+1)(2n+1)}{6}) +3(\frac{n(n+1)}{2})=\\3(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})=3(\frac{n(n+1)(n+2)}{3})$

3. zmudz

yes

4. amoodarya

I almost solve it and the last step is ...?

5. zmudz

cancel out the 3s and get n(n+1)(n+2) then expand. thank you!

6. amoodarya

I did not wrote the simplification steps complete hope you do it