Question

The value of \[ \frac{{}^{100}C_{50}}{51} + \frac{{}^{100}C_{51}}{52} + \cdots + \frac{{}^{100}C_{100}}{101} \] is:

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

Hint:

Whenever a binomial term is divided by \(r+1\), try converting it into a higher combination using \(\displaystyle \frac{{}^nC_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1}\).

Answer 1

The Correct Option is C

Concept:
We use the identity: \[ \frac{{}^nC_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1} \] and the binomial theorem: \[ \sum_{k=0}^{n} {}^nC_k = 2^n \]
Step 1: Rewrite each term using the identity. \[ \frac{{}^{100}C_r}{r+1} = \frac{{}^{101}C_{r+1}}{101} \] Thus the given sum becomes: \[ \frac{1}{101}\left({}^{101}C_{51} + {}^{101}C_{52} + \cdots + {}^{101}C_{101}\right) \]
Step 2: Use symmetry of binomial coefficients. \[ \sum_{k=0}^{101} {}^{101}C_k = 2^{101} \] Also, \[ {}^{101}C_0 + {}^{101}C_1 + \cdots + {}^{101}C_{50} = 2^{100} \] (by symmetry of terms about the middle)
Step 3: Evaluate the required sum. \[ {}^{101}C_{51} + \cdots + {}^{101}C_{101} = 2^{101} - 2^{100} \] \[ = 2^{100} \]
Step 4: Final value. \[ \text{Required sum} = \frac{2^{100}}{101} \]