2024 DMC Mock 10 Problems/Problem 10

Notice that $a_n-a_{n-1}=2(a_{n-1}-a_{n-2})$ . Since $a_2-a_1=1$ , we get $a_{100}=(a_{100}-a_{99})+(a_{99}-a_{98})+\dots+(a_2-a_1)=2^{98}+2^{97}+\dots+1 = 2^{99}-1$ so $a_{100}+1=2^{99}$ . Therefore the answer is $\boxed{99}$ .