The problem is:
How many subsets of {${1,2,...,n}$} have an even sum of their elements?
Please take a look at my solution since this is my answer from the exam and I think to appeal on this question, I'd like to hear your thoughts if I can appeal this.
I decided to solve this problem by recursive formula, my solution:
Let's define:
$a_{n}$ - The number of subsets of {${1,2,...,n}$} that have an even sum of their elements.
Let's split into two different cases:
$*$1. $n$ is odd, therefore in this case we got: $2^{n-1}-a_{n-1}$
2. $n$ is even, therefore in this case we got: $a_{n-1}$
$*$ EXPLANATION: $2^{n-1}$ is the number of subsets of {${1,2,...,n}$}.
$a_{n-1}$ is the number of subsets of {${1,2,...,n}$} that have an even sum of their elements like I defined. Therefore $2^{n-1}-a_{n-1}$ is the number of subsets of {1,2,...,n} that have an odd sum of their elements.
From summarize the cases we got: $a_{n} = 2^{n-1}-a_{n-1}+a_{n-1}$, Therefore $a_{n} = 2^{n-1}$
The answer to the main question is $a_{11} = 2^{11-1} = 2^{10}$.
The comment I got from my teacher about this answer is about the split into the cases, he wrote that $n$ don't have to be a part of the group and he is completely right.
In fact, my final answer was right, but my split into the two cases is completely wrong and I received 0% points for this answer. I would like to appeal this question in claiming that I meant other cases separation and it was only a formulation mistake.
I'll be glad if someone can tell me if it makes sense, and if it is, how to formulate it right to gain some points.