Sum of odd numbers is odd if each of the natural numbers is odd

Question

Prove that the sum of an odd number of natural numbers is odd if each of the natural numbers is odd.

Here's what I tried already but it didn't work:

$\sum_{i=0}^n i = 2n-1$ but when I use induction, I end up with $2(n+1)-1$ which would be an even number. Any suggestions?

Answer 1

What you have is $$ \sum_{j=1}^{2k+1}(2n_j+1)=2\left(\sum_{j=1}^{2k+1}n_j\right) +2k+1=2\left [\left (\sum_{j=1}^{2k+1}n_j \right)+k\right]+1. $$

Answer 2

Hint:-

$a$ is odd $\iff a\equiv1\pmod 2$.

Answer 3

The sum of an odd number of odd numbers is the sum of several pairs of odd numbers plus an odd number.

The sum of a pair of odd numbers is even. [(2k + 1) + (2j + 1) = 2(j + k + 1)]

The sum of any number of even numbers is even. [2k + 2j + .... + 2l = 2(k+j+...l)]

The sum of any even number plus an odd number is odd. [2k + 2j + 1 = 2(k + j) + 1]

The sum of an odd number of odd numbers is the sum of several pairs of odd number (the sum of each pair is even; the sum of the several pairs is even) plus an odd number (which is odd, so when added to the even sum of several pairs of odd numbers yields an odd number) so the total is odd.

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Or $\sum_{i=1}^{2k+1} (2j_i + 1) = 2(\sum_{i=1}^{2k+1}k) + \sum_{i=1}^{2k+1}1 = 2(\sum_{i=1}^{2k+1}k) + (2k + 1) = 2[(\sum_{i=1}^{2k+1}k) + k] + 1$ which is odd.