Question

The value of the limit \[ \lim_{n \to \infty} \sum_{k=0}^{n} \binom{2n}{k} \frac{1}{4^n} \] is equal to:

• A. 1
• B. \(\frac{1}{2}\)
• C. 0
• D. \(\frac{1}{4}\)

Hint:

In binomial expansions, the first half of coefficients add up to half of the total sum when \( n \) is even or large.

Answer 1

The Correct Option is B

Step 1: Understanding the expression.
We know that \[ \sum_{k=0}^{2n} \binom{2n}{k} \frac{1}{4^n} = \left(\frac{1}{2} + \frac{1}{2}\right)^{2n} = 1 \] But the question sums only up to \( k = n \).
Step 2: Using symmetry of the binomial coefficients.
Because of the symmetric nature of binomial coefficients, \[ \sum_{k=0}^{n} \binom{2n}{k} = \frac{1}{2} \times \sum_{k=0}^{2n} \binom{2n}{k} \] Thus, \[ \sum_{k=0}^{n} \binom{2n}{k} = \frac{1}{2} \times 2^{2n} = 2^{2n - 1} \]
Step 3: Substitute into the given expression.
\[ \lim_{n \to \infty} \frac{2^{2n - 1}}{4^n} = \frac{1}{2} \] Final Answer: \[ \boxed{\frac{1}{2}} \]