Lets say we have $n$ coins and probability of Coin $i$ to fall heads is $f(i)$ . Find the probability of getting an even number of heads when all the $n$ coins are tossed.
$f(i) = \frac {1} {2i + 3}$
Here $n$ is large, of the order of 1e5, so an efficient approach is required.
I tried to analyze the brute force case but that would be too much i.e. If I calculate for 2 success, 4, 6..., it might take years to run.
Then I thought of applying Linearity of Expectation somehow but couldn't come up with anything which could help.
Hint
There is a simple recurrence equality which is provable by induction.
Deduce one from the fact that (denoting $H_n$ the $n$-th instance of your problem) $$P(H_n) = p_n\cdot P(\neg H_{n-1})+ (1-p_n) \cdot P(H_{n-1})$$
This allows you to iteratively compute $P(H_n)$. Simplify this to a closed form.