Question

The value of \( \frac{100 C_{50}}{51} + \frac{100 C_{51}}{52} + \dots + \frac{100 C_{100}}{101} \) is :

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

Hint:

If the upper index of the sum of binomial coefficients is odd (like 101), the sum of the first half of terms is exactly equal to the sum of the second half.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use the identity \(\frac{^nC_r}{r+1} = \frac{^{n+1}C_{r+1}}{n+1}\). This converts the sum into a sum of binomial coefficients of the next order.
Step 2: Key Formula or Approach:
\(\sum_{r=k}^n \frac{^nC_r}{r+1} = \frac{1}{n+1} \sum_{r=k}^n ^{n+1}C_{r+1}\).
Step 3: Detailed Explanation:
The expression is: \[ \frac{1}{101} [ ^{101}C_{51} + ^{101}C_{52} + \dots + ^{101}C_{101} ] \] Sum of all coefficients for \(n=101\) is \(2^{101}\). Due to symmetry \(^{101}C_r = ^{101}C_{101-r}\), the sum of the second half of the coefficients (from 51 to 101) is exactly half of the total sum. Sum \( = \frac{1}{2} (2^{101}) = 2^{100}\). Final Value \( = \frac{2^{100}}{101}\).
Step 4: Final Answer:
(4) \( \frac{2^{100}}{101} \).