Solving the Summation: From i0 to n-2, n - 1 - i

In this article, we explore a specific summation problem often encountered in mathematical contexts. Specifically, we aim to solve the sum from (i0) to (n-2)

∑i0to(n-2)(n-1-i)

Reformulating the Problem in LaTeX

To make the problem more clear, let's reformulate it using LaTeX:

∑i0to(n-2)(n-1-i)

Mathematically, this can be rewritten as:

S∑i0to(n-2)(n-1-i)

By shifting the index (i), we can simplify the expression as follows:

S∑i1to(n-1)(n-i)

Breaking Down the Summation

The problem can be broken down into two parts, each significantly simpler to solve:

The summation of (n) over the range from 1 to (n-1) The summation of (i) over the range from 1 to (n-1)

Let's evaluate each part separately:

Part 1: Summation of (n)

The first part is:

N∑n1to(n-1)n

This simplifies to:

Nn×(n-1)

Part 2: Summation of (i)

The second part is:

I∑n1to(n-1)∑i

The sum of the first (n-1) integers can be found using the formula:

∑1to(n-1)imm-12×(n-1)

Where (m n-1):

In-12×(n-1)

Final Calculation

By combining both parts, we get:

SN-In×(n-1)-n-12×(n-1)

Which simplifies to:

S2nmi-3-12×(n-1)n-12×(2n-n-1)

Which further simplifies to:

Sn-12×(n-1)n(n-1)2-(n-1)(n-1)2n(n-1)-(n-1)(n-1)2n-12

Therefore, the result of the summation is:

Sn-12

Conclusion

This solution shows us that the summation of (n - 1 - i) from (i0) to (n-2) can be simplified significantly, and the result is intriguingly symmetrical. By understanding the breakdown and simplification, we can tackle similar problems more effectively.

References

[1] Wikipedia - Summation

[2] MathWorld - Sum