Find polynomial f(n) such that for all integers n>=1, we have 3(1*2+2*3+...+n(n+1))=f(n). Write f(n) as a polynomial with terms in descending order or n.

Find polynomial f(n) such that for all integers n>=1, we have 3(1*2+2*3+...+n(n+1))=f(n). Write f(n) as a polynomial with terms in descending order or n.

\(\begin{array}{|rcll|} \hline f(n) &=& 3\cdot[~ 1\cdot 2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 + \dots +n(n+1)~] \\ f(n) &=& 3\cdot[~ 1\cdot (1+1) + 2\cdot (2+1) + 3\cdot (3+1) + 4\cdot (4+1) + \dots +n(n+1)~] \\ f(n) &=& 3\cdot[~ (1^2+1) + (2^2+2) + (3^3+3) + (4^2+4) + \dots +(n^2+n)~] \\ f(n) &=& 3\cdot[~ (1+2+3+4+ \dots +n) + (1^2+2^2+3^3+4^2 + \dots +n^2)~] \\\\ && (1+2+3+4+ \dots +n) = \frac{(1+n)\cdot n}{2}\\ && (1^2+2^2+3^3+4^2 + \dots +n^2)= \frac{n\cdot(n+1)\cdot(2n+1)}{6} \\\\ f(n) &=& 3\cdot[~ \frac{(1+n)\cdot n}{2} + \frac{n\cdot(n+1)\cdot(2n+1)}{6}~] \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot[~ \frac{1}{2} + \frac{(2n+1)}{6}~] \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot[~ \frac{3}{6} + \frac{(2n+1)}{6}~] \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot( \frac{3+2n+1}{6} ) \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot( \frac{2n+4}{6} ) \\ f(n) &=& 3\cdot n \cdot(n+1)\cdot 2\cdot( \frac{n+2}{6} ) \\ \mathbf{f(n) }&\mathbf{=} &\mathbf{ n \cdot (n+1)\cdot (n+2)} \\ \hline \end{array}\)