Question

Let \( a_0 = 0 \) and define \( a_n = \frac{1}{2} (1 + a_{n-1}) \) for all positive integers \( n \geq 1 \). The least value of \( n \) for which \( |1 - a_n|<\frac{1}{2^{10}} \) is ______.

Hint:

When dealing with recurrence relations that approach a limit, find the general form of the sequence and solve for the required condition.

Answer 1

Correct Answer: 11

We are given the recurrence relation \( a_n = \frac{1}{2} (1 + a_{n-1}) \) with the initial condition \( a_0 = 0 \). We need to find the least value of \( n \) for which \( |1 - a_n|<\frac{1}{2^{10}} \). Step 1: Observing the recurrence relation, we see that the sequence \( a_n \) converges to 1. We can express \( a_n \) as: \[ a_n = 1 - \frac{1}{2^n}. \] This is a standard result for such recurrence relations, where \( a_n \) converges to 1. Step 2: We need to find \( n \) such that: \[ |1 - a_n| = \left| 1 - \left( 1 - \frac{1}{2^n} \right) \right| = \frac{1}{2^n}<\frac{1}{2^{10}}. \] This inequality simplifies to: \[ \frac{1}{2^n}<\frac{1}{2^{10}}. \] Step 3: Solving for \( n \), we get: \[ n \geq 11. \] Thus, the least value of \( n \) is 11.