The problem you provided is asking to prove the following equality for any positive integer $n$:

$0 \cdot n + 1 \cdot (n-1) + 2 \cdot (n-2) + \cdots + (n-1) \cdot 1 + n \cdot 0 = \frac{(n-1)n(n+1)}{6}$

Approach

This sum represents a telescoping series of products of consecutive integers. Here's a step-by-step approach to prove the equality:

1. Expression of the Sum: The sum on the left-hand side is:

$S = 0 \cdot n + 1 \cdot (n-1) + 2 \cdot (n-2) + \cdots + (n-1) \cdot 1 + n \cdot 0$

which can be written as:

$S = \sum_{k=1}^{n-1} k \cdot (n-k)$

2. Simplify Each Term: Each term $k \cdot (n-k)$ can be expanded as:

$k \cdot (n-k) = k \cdot n - k^2$

Thus, the sum becomes:

$S = \sum_{k=1}^{n-1} (k \cdot n - k^2)$

This can be split into two separate sums:

$S = n \cdot \sum_{k=1}^{n-1} k - \sum_{k=1}^{n-1} k^2$

3. Evaluate the Two Sums:
   • The first sum is the sum of the first $n-1$ positive integers:

$\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}$

• The second sum is the sum of squares of the first $n-1$ integers. Using the formula for the sum of squares:

$\sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}$

4. Substitute Back: Now substitute these results back into the expression for $S$:

$S = n \cdot \frac{(n-1)n}{2} - \frac{(n-1)n(2n-1)}{6}$

Factor out $\frac{(n-1)n}{6}$:

$S = \frac{(n-1)n}{6} \left( 3n - (2n-1) \right)$

Simplifying the terms inside the parentheses:

$S = \frac{(n-1)n}{6} \left( 3n - 2n + 1 \right) = \frac{(n-1)n(n+1)}{6}$

Thus, the original sum is equal to the right-hand side of the equation, which completes the proof:

$S = \frac{(n-1)n(n+1)}{6}$

Conclusion

The equality has been proven as requested.

Let me know if you would like further details or clarifications!

Related Questions:

1. How do you calculate sums of series with products like this for other patterns?
2. What are the general forms of sum of squares and sum of cubes formulas?
3. Can this telescoping technique be applied to other series?
4. What other proof techniques can be used for such summations?
5. How can induction be applied to this problem?

Tip:

When dealing with sums involving products, try to simplify each product term first and look for recognizable patterns (like sums of integers or squares).