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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).
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).

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amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1do you mean f(n)=\[3(1*2+2*3+3*4+...+n(n+1))\]?

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1\[3(1*2+2*3+3*4+...+n(n+1))=3\sum_{1}^{n}i(i+1)=\\3\sum_{1}^{n}(i^2+i)=\\3\sum_{1}^{n}(i^2)+3\sum_{1}^{n}(i)=\\3(\frac{n(n+1)(2n+1)}{6}) +3(\frac{n(n+1)}{2})=\\3(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})=3(\frac{n(n+1)(n+2)}{3})\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1I almost solve it and the last step is ...?

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0cancel out the 3s and get n(n+1)(n+2) then expand. thank you!

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1I did not wrote the simplification steps complete hope you do it
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