Let $ A $ be the set formed by the $ n $ terms of the arithmetic progression given by $$ \begin {cases} a_1 = 1 \\ r = 1 \\ \end {cases} $$Knowing that $ med (A) $ does not belong to $ A $, how many subsets of $ A $ contain odd numbers?
Note: $ med (A) $ represents the statistical median of the values of $ A $
Solution: The progression is just $1, 2, \dots, n,$ which does not contain the median $(1+n)/2$ iff $n$ is even. Letting $n = 2k,$ the possible subsets are subsets of $\{1, 3, \dots, 2k-1\}$ for a total of $2^k = 2^{n/2}$ possible subsets.
Correct?
There are $2^{2k}$ subsets of $A$, and there are $2^k$ subsets of $A$ which contain only even numbers (and hence which do not contain odd numbers).
It follows that there are $2^{2k}-2^k = 2^k(2^k-1)$ subsets of $A$ which contain at least one odd number.
(Do you now understand where your solution went wrong? You calculated the number of subsets $A$ which contain only odd numbers.)