# Prove Summation Equation: (2^n-1)n for Any Integer n

• tiny-tim
In summary, the given equation \sum_{0\,\leq\,m\,< n/2}\,(-1)^m(n - 2m)^n\,^nC_m\ =\ 2^{n-1}\,n!, can be proved by using a geometric-cum-combinatoric approach. It is also listed as an exercise in the book "A = B" by Marko Petkovsek, Herbert Wilf, and Doron Zeilberger.
tiny-tim
Homework Helper
Prove, for any integer n:

$$\sum_{0\,\leq\,m\,< n/2}\,(-1)^m(n - 2m)^n\,^nC_m\ =\ 2^{n-1}\,n!$$

for example, 77 - 577 + 3721 - 1735

= 823543 - 546875 + 45927 - 35 = 304560 = 64 times 5040

I'm guessing there is a brute force way to prove this as well as a clever combinatoric argument, and you're hoping we only find the first, so you can dazzle us with the second?

Gokul43201 said:
I'm guessing there is a brute force way to prove this as well as a clever combinatoric argument, and you're hoping we only find the first, so you can dazzle us with the second?

Hi Gokul!

Sort of … I accidentally found a geometric-cum-combinatoric proof of this while looking at a homework thread,

but I couldn't help thinking that there must be some way of solving this just by looking at it and coming up with a solution …

but no ordinary technique comes to mind since the exponand (is that the right word? ) keeps changing.

I was hoping somebody knew a finding-the-solution technique (maybe for a simpler problem), rather than an already-knowing-what the-solution-is technique!

## What is the summation equation (2^n-1)n?

The summation equation (2^n-1)n represents the sum of the first n terms of a geometric sequence where each term is equal to 2^n-1.

## How do you prove the summation equation (2^n-1)n for any integer n?

To prove the summation equation (2^n-1)n, we can use mathematical induction. We first show that the equation holds for n=1, then assume it holds for some arbitrary integer k, and finally prove that it also holds for k+1. This will demonstrate that the equation holds for all positive integers.

## What is mathematical induction?

Mathematical induction is a proof technique used to prove that a statement is true for all positive integers. It involves showing that the statement holds for the first integer (often n=1), assuming it holds for an arbitrary integer k, and then proving that it also holds for k+1. This process can be repeated infinitely to show that the statement holds for all positive integers.

## What are the steps to prove the summation equation (2^n-1)n using mathematical induction?

Step 1: Show that the equation holds for n=1.Step 2: Assume the equation holds for some arbitrary integer k.Step 3: Prove that the equation also holds for k+1.Step 4: Conclude that the equation holds for all positive integers by the principle of mathematical induction.

## Why is it important to prove the summation equation (2^n-1)n?

Proving the summation equation (2^n-1)n is important because it allows us to accurately calculate the sum of a geometric sequence with an infinite number of terms. This equation is also used in various mathematical and scientific fields, making it a valuable tool for solving problems and making calculations.

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