Proof there is no way to chose signs to make sequential sum $1+2+3+\cdots+10$ even

Question

I've figured that for the sum

$$1+2+3+4+5+6+7+8+9+10=55$$

There is no way to chose the signs of the numbers to get an even sum.

I'm really struggling to prove this and would appreciate some assistance.

Also for interests sake, could this be proved generally (no way to chose signs of any sequential sum that equals an uneven number to get an even one?)

Answer 1

The parity of a number does not change when you change its sign.

Answer 2

A sum of integers $a_1+\cdots+a_n$ is odd if and only if an odd number of terms $a_i$ are odd (that was a lot of odds for one sentence). Since changing sign doesn't change parity, the parity of the sum is not changed either.

Answer 3

No matter what the signs are, you have five even numbers (whose sum is even) and an odd number (five) of odd numbers. That's all you need to know to conclude that the sum is odd.

Answer 4

When any specific term (call it $a$) has its sign reversed, what you're doing is subtracting $a$ twice from the sum $S$, so the sum becomes $S - 2a$. This extends to any number of terms so you are always subtracting an even number, which never changes the parity of the sum.

EDIT: I just noticed @lulu's comment, which covered this.

Answer 5

Let $(a_n)_n$ be the terms that we sum. You are asking if $\sum_{i=1}^{n} (-1)^{b_i}*a_i \equiv \sum_{i=1}^{n} a_i \mod 2$ for some $(b_n)_n \in \{0, 1\}$.

We use $a*b \mod c \equiv (a \mod c) * (b \mod c) \mod c$:

$\sum_{i=1}^{n} (-1)^{b_i}*a_i \equiv \sum_{i=1}^{n} ((-1)^{b_i} \mod 2) * (a_i \mod 2)\mod 2$. Now using the fact that $(-1)^m \equiv 1 \mod 2, \forall m \in N$ we conclude:

$\sum_{i=1}^{n} (-1)^{b_i}*a_i \equiv \sum_{i=1}^{n} 1 * a_i\mod 2$