Prove that for any n > 1, the sum of n odd numbers in a row(for example: 5 ,7, 9) cannot be a prime number.?

Question

If you say to use a specific method, please explain the steps.
Thanks in advance!

Answer 1

Here's a qualitative proof.

Ok, n can be odd or even.

If n is even, you're pretty much done as the sum of any even number of odd numbers is an even number, thus is not prime.

If n is odd (and > 1), then the minimum for n is 3.
The sum of the series (x-2), x, (x+2) will not be prime because it will be divisible by 3 and would in fact be 3x (average of x)

For every odd number greater than 3, you can expand the series out from the middle number into x-4, x-2, x, x+2, x+4 and so forth, continually averaging x.

Thus, since the series of n numbers will always be divisible by x, the middle number, and n, the number of odd numbers in a row, it is not prime.

Hope that helps!

Answer 2

Let's use an example to illustrate:

ex: n = 15.
The sum of 15 numbers starting at x expands to:
x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12, x + 14, x +1 6, x + 18, x + 20, x + 22, x + 24, x + 26, x + 28

Simplify that by splitting out the x's and the numbers:
15x + (0 + 2 + 4...+ 28)
15x + 2(0 + 1 + 2... + 14)

Note that 15x = nx, and 2(0 + 1 + 2... + 14) can be rewritten as a summation of the numbers from 0 to n-1...
nx + 2(Σ 0 -> n-1)

The sum of n consecutive numbers is (n * (n + 1))/2. In this case, 0 is ignored, so we have n-1 consecutive numbers:
2((n-1) * n)/2. The 2's cancel, which leaves us with n-1 * n, or n^2 - n.

Giving us:
nx + n^2 - n
n^2 + (x -1)n
Refactor:
n(n + x -1)

Therefore, no matter what n is, your answer is a multiple of n - thus, it can never be a prime number unless n is 1.