Math Help - Standardized Test Question

1. Standardized Test Question

What is the most efficient approach to this question?
I am assuming there is some geometric series shortcut which would give me a shortcut. Or is it just a matter of looking at the properties of the final number without actually taking the time to figure it out.

2. Of course, they don't expect you to calculate that, but take advantage of a cool property of this function.

As defined $h(n)=2*4*6*8*...*2k$, where $2k=n$

so $h(100)=2*4*6*8*...100$
Factorize
$h(100)=2(1*2*3*4*...*50)$
$h(100)=2*50!$

Now $h(100)$ is divisible by all the primes below 50, but when we add 1
$h(100)+1=2*50!+1$

It is no longer divisible by any number below 50.