What is the Value of the Summation n-13?

In this article, we delve into the mathematical concept of the summation of cubes, specifically the expression (sum_{n1}^{N} (n-1)^3). We will explore the mathematical derivation, proof, and practical application of this expression. Understanding this concept is essential for students and professionals in various fields, including mathematics, computer science, and engineering.

The Summation Expression and Its Simplification

The original expression given is:

[sum_{n1}^{N} (n-1)^3]

By substituting (k n-1), we can rewrite the summation as:

[sum_{n1}^{N} (n-1)^3 sum_{k0}^{N-1} k^3]

This transformation simplifies the range of the summation, making it easier to analyze and solve.

Deriving the Sum of Cubes

The formula for the sum of the first (m) cubes is given by:

[sum_{k1}^{m} k^3 left(frac{m(m 1)}{2}right)^2]

Applying this formula to our case where (m N-1) yields:

[sum_{k0}^{N-1} k^3 sum_{k1}^{N-1} k^3 left(frac{(N-1)N}{2}right)^2]

Thus, the value of the summation (sum_{n1}^{N} (n-1)^3) is:

[left(frac{(N-1)N}{2}right)^2]

Enhanced Mathematical Proof

To further validate our result, we can use a first-order difference equation approach. Given:

[S_N sum_{n0}^{N} n^3]

We find:

[S_N - S_{N-1} N^3]

This is a first-order difference equation, and we solve it with the trial solution:

[S_N aN^4 bN^3 cN^2 dN e]

Plug this into the difference equation and equate like terms:

[4a 1] [-6a 3b 0] [4a - 3b 2c 0] [-a b - c 0]

Solving these equations gives:

[a frac{1}{4}] [b frac{1}{2}] [c frac{1}{4}] [d 0] [e 0]

Therefore:

[S_N frac{N^4}{4} frac{N^3}{2} frac{N^2}{4}]

Which simplifies to:

[S_N frac{N^2(N 1)^2}{4}]

Substituting (N-1) for (N) gives:

[S_{N-1} frac{(N-1)^2N^2}{4}]

Practical Application and Conclusion

This mathematical derivation is not just theoretical. It has numerous practical applications, particularly in computer science and engineering. For instance, in algorithms and data structures, understanding the sum of cubes can help optimize computational efficiency.

For a specific value of (N), you can substitute it into the formula to find the numerical result. For example, if (N 5), the value of the summation is:

[left(frac{(5-1)5}{2}right)^2 100]

In conclusion, the summation of cubes, specifically the expression (sum_{n1}^{N} (n-1)^3), is a fundamental concept in mathematics. Understanding its derivation and application can significantly enhance problem-solving skills in various fields.