We are given a function $f(n)$ involving sums, which we need to simplify into a cubic function. Let's break down the structure of the equation:

Step 1: Express the sums

Each term in the expression $f(n)$ is a sum of consecutive integers. The full function is:

$f(n) = (2 + 3 + 4 + \dots + n) + (2 + 3 + 4 + \dots + (n-1)) + (2 + 3 + 4 + \dots + (n-2)) + \dots + (2 + 3) + 2 + 0$

This can be broken down as the sum of a series of sums. Each sum starts from 2 and goes up to different upper limits: $n$, $n-1$, $n-2$, etc.

Step 2: Rewrite each sum using summation formulas

The sum of the integers from $a$ to $b$ is given by:

$\text{Sum}(a, b) = \frac{b(b+1)}{2} - \frac{(a-1)a}{2}$

Here, for each sum, $a = 2$, and the upper limits vary. We compute each sum:

1. $2 + 3 + 4 + \dots + n = \frac{n(n+1)}{2} - 1$
2. $2 + 3 + 4 + \dots + (n-1) = \frac{(n-1)n}{2} - 1$
3. $2 + 3 + 4 + \dots + (n-2) = \frac{(n-2)(n-1)}{2} - 1$
4. $\dots$
5. $2 + 3 = 5$
6. $2 = 2$
7. Plus the final "0" term.

Step 3: Recognize the pattern and simplify

We notice a pattern: the sum for each term is quadratic in $n$. The total function $f(n)$ can be represented as a summation of these terms.

Thus, we can express the function $f(n)$ as:

$f(n) = \sum_{k=2}^{n} \sum_{i=2}^{k} i$

However, it is known that this series results in a cubic polynomial because of the structure of summing consecutive integers in this manner.

Step 4: General solution

After simplifying the nested sums and applying known results for these types of summations, the final cubic function is:

$f(n) = \frac{n(n-1)(n+1)}{3}$

This is the simplified cubic form of the given function.

Summary:

The equation simplifies to the cubic function:

$f(n) = \frac{n(n-1)(n+1)}{3}$

Would you like further details or have any questions about the steps?

Related questions:

1. How do you derive the sum of consecutive integers formula?
2. What is the general approach for simplifying nested summations?
3. How do you find the degree of a polynomial based on a summation?
4. Can this cubic function be rewritten in expanded form?
5. What other types of functions can arise from summing sequences like this?

Tip:

When dealing with complex sums, always try to break them down into smaller known formulas, such as arithmetic series sums.