We can first expand the sum inside the parentheses to get
\[1\cdot1 + 2\cdot2 + \dots + n\cdot n = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}.\]
Thus, we have
\[6\left(1\cdot1 + 2\cdot2 + \dots + n\cdot n\right) = 6\cdot \frac{n(n+1)(2n+1)}{6} = n(n+1)(2n+1).\]
Therefore, the polynomial we are looking for is
\[f(n) = n(n+1)(2n+1) = 2n^3 + 3n^2 + n.\]
Find polynomial f(n) such that for all integers n \geq 1, we have
6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right) = f(n).\]
Write f(n) as a polynomial with terms in descending order of n.
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