A sequence is defined by a_1 = 1 and a_n = 2a_{n-1} + 1 for n >= 2. Find the smallest n such that a_n > 1000.
Correct: C
The recurrence relation is a_n = 2a_{n-1} + 1.
We can rewrite this as a_n + 1 = 2a_{n-1} + 2, which simplifies to a_n + 1 = 2(a_{n-1} + 1).
Let b_n = a_n + 1. Then the recurrence relation becomes b_n = 2b_{n-1}.
This means that the sequence b_n is a geometric progression with a common ratio of 2.
The first term b_1 = a_1 + 1 = 1 + 1 = 2.
So, b_n = b_1 * 2^(n-1) = 2 * 2^(n-1) = 2^n.
Since b_n = a_n + 1, we have a_n = b_n - 1 = 2^n - 1.
We need to find the smallest n such that a_n > 1000.
So, 2^n - 1 > 1000.
2^n > 1001.
Let's check powers of 2:
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = 1024
The smallest integer n for which 2^n > 1001 is n = 10.