Analyzing the Recursive Sequence

We are given a sequence defined by $a_0 = 0$ and the recurrence relation $a_n = \frac{1}{2}(1 + a_{n-1})$ for $n \ge 1$. We need to find the least integer $n$ such that $|1 - a_n| < \frac{1}{2^{10}}$.

Calculating Sequence Terms

Let's compute the first few terms to find a pattern:

• $a_0 = 0$
• $a_1 = \frac{1}{2}(1 + a_0) = \frac{1}{2}(1 + 0) = \frac{1}{2}$
• $a_2 = \frac{1}{2}(1 + a_1) = \frac{1}{2}(1 + \frac{1}{2}) = \frac{1}{2}(\frac{3}{2}) = \frac{3}{4}$
• $a_3 = \frac{1}{2}(1 + a_2) = \frac{1}{2}(1 + \frac{3}{4}) = \frac{1}{2}(\frac{7}{4}) = \frac{7}{8}$

Deriving the Closed-Form Formula

Observe the pattern:

• $a_1 = \frac{1}{2} = 1 - \frac{1}{2}$
• $a_2 = \frac{3}{4} = 1 - \frac{1}{4}$
• $a_3 = \frac{7}{8} = 1 - \frac{1}{8}$

This suggests the formula $a_n = 1 - \frac{1}{2^n}$ for $n \ge 1$. Let's verify this using the recurrence relation. If $a_{n-1} = 1 - \frac{1}{2^{n-1}}$, then $a_n = \frac{1}{2}(1 + a_{n-1}) = \frac{1}{2}(1 + (1 - \frac{1}{2^{n-1}})) = \frac{1}{2}(2 - \frac{1}{2^{n-1}}) = 1 - \frac{1}{2} \cdot \frac{1}{2^{n-1}} = 1 - \frac{1}{2^n}$. This formula holds for $n \ge 1$. It also holds for $n=0$ since $a_0 = 1 - \frac{1}{2^0} = 1 - 1 = 0$.

Therefore, the closed-form formula for the sequence is $a_n = 1 - \frac{1}{2^n}$ for all $n \ge 0$.

Solving the Inequality

We need to find the least $n$ such that $|1 - a_n| < \frac{1}{2^{10}}$.

Substitute the formula for $a_n$: $|1 - (1 - \frac{1}{2^n})| < \frac{1}{2^{10}}$ $|\frac{1}{2^n}| < \frac{1}{2^{10}}$

Since $n$ is a positive integer, $\frac{1}{2^n}$ is positive. The inequality simplifies to:

For this inequality to hold, the exponent $n$ must be greater than the exponent $10$.

Determining the Least Integer Value

The least integer value of $n$ that satisfies the condition $n > 10$ is $n = 11$. This value is a positive integer, consistent with the problem statement ($n \ge 1$).