First, let's expand the sum to see a pattern:
\[6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right) = 6\left( 1 + 4 + \dots + n^2 \right)\]
\[= 6\left(1 + 4 + 9 + \dots + n^2\right)\]
Now, notice that the sum of squares from $1^2$ to $n^2$ can be represented by the formula for the sum of the first $n$ perfect squares:
\[f(n) = 6\left(1 + 4 + 9 + \dots + n^2\right) = 6\left(\frac{n(n + 1)(2n + 1)}{6}\right)\]
Simplifying, we get:
\[f(n) = (n)(n + 1)(2n + 1)\]
Thus, the polynomial $f(n)$ is:
\[f(n) = 2n^3 + 3n^2 + n\]
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 $n.$
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