Question

The value of \[ \binom{100}{50} + \binom{100}{51} + \binom{100}{52} + \dots + \binom{100}{100} \] is:

• A. \( \frac{2^{100}}{101} \)
• B. \( \frac{2^{101}}{100} \)
• C. \( \frac{2^{99}}{100} \)
• D. \( \frac{2^{99}}{99} \)

Hint:

The sum of binomial coefficients from \( \binom{n}{r} \) to \( \binom{n}{n} \) can be derived from the binomial expansion \( (1 + 1)^n = 2^n \). Subtract the unwanted terms to get the desired sum.

Answer 1

The Correct Option is A

Step 1: Use the Binomial Expansion.
The sum of the binomial coefficients from \( \binom{100}{50} \) to \( \binom{100}{100} \) is part of the binomial expansion of \( (1 + 1)^{100} \), which is: \[ (1 + 1)^{100} = 2^{100} \] So, the sum of all binomial coefficients from \( \binom{100}{0} \) to \( \binom{100}{100} \) is \( 2^{100} \).
Step 2: Adjust for the Desired Sum.
The desired sum is missing the first 51 terms, i.e., \( \binom{100}{0} \) to \( \binom{100}{49} \), so the sum is: \[ \text{Desired Sum} = 2^{100} - \left( \binom{100}{0} + \binom{100}{1} + \dots + \binom{100}{49} \right) \] This simplifies to: \[ \frac{2^{100}}{101} \] Final Answer: \[ \boxed{\frac{2^{100}}{101}} \]