Question

Find polynomial f(n) such that for all integers n is greater than or equal to 1 we have 3t( 1*2 + 2*3 +...+ n(n+1) ) = f(n). Write f(n) as a polynomial with terms in descending order of n

Answer 1

$1\cdot2+2\cdot3+\cdots+(n-1)\cdot n+n\cdot(n+1)=\displaystyle\sum_{k=1}^nk(k+1)=\sum_{k=1}^n(k^2+k)$
$=\dfrac{n(n+1)(2n+1)}6+\dfrac{n(n+1)}2$
$=\dfrac{n(n+1)(2n+1)+3n(n+1)}6$
$=\dfrac{2n^3+6n^2+4n}6$
$=\dfrac13n^3+n^2+\dfrac23n$