SOLUTION: The sum 6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right) is equal to a polynomial f(n) for all n \ge 1. Write f(n) as a polynomial with terms in descending order of n.

Algebra ->  Sequences-and-series -> SOLUTION: The sum 6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right) is equal to a polynomial f(n) for all n \ge 1. Write f(n) as a polynomial with terms in descending order of n.      Log On


   



Question 1207587: The sum
6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right)
is equal to a polynomial f(n) for all n \ge 1.
Write f(n) as a polynomial with terms in descending order of n.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
I finally figured out that the above, which was gobbledygook to me, 
means this:

The sum
6%281%2A1%2B2%2A2%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Bn%2An%29
is equal to a polynomial f(n) for all n%3E=1.
Write f(n) as a polynomial with terms in descending order of n.
It is well-known that

1%5E2%2B2%5E2%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Bn%5E2%22%22=%22%22%28n%28n%2B1%29%282n%2B1%29%29%2F6

so

6%281%2A1%2B2%2A2%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Bn%2An%29%22%22=%22%226%281%5E2%2B2%5E2%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Bn%5E2%29%22%22=%22%226%2Aexpr%28%28n%28n%2B1%29%282n%2B1%29%29%2F6%29%22%22=%22%22
cross%286%29%2Aexpr%28%28n%28n%2B1%29%282n%2B1%29%29%2Fcross%286%29%29%22%22=%22%22n%28n%2B1%29%282n%2B1%29%22%22=%22%22n%282n%5E2%2B3n%2B1%29%22%22=%22%222n%5E3%2B3n%5E2%2Bn

So

%22f%28n%29%22%22%22=%22%222n%5E3%2B3n%5E2%2Bn

Edwin