Question

Let \( \alpha = \sum_{k=0}^{5} {^{10}C_{2k}} \) and \( \beta = \sum_{k=0}^{4} {^{10}C_{2k+1}} \). Then \( \alpha - \beta \) is equal to

• A. \(32 \)
• B. \(64 \)
• C. \(128 \)
• D. \(256 \)
• E. \(0 \)

Hint:

Sum of even and odd binomial coefficients are equal: each is \(2^{n-1}\).

Answer 1

Answer: (E)

Concept: \[ (1+1)^n = \sum_{k=0}^{n} {^nC_k}, \quad (1-1)^n = \sum_{k=0}^{n} (-1)^k {^nC_k} \] From identities: \[ \text{Sum of even terms} = \text{Sum of odd terms} = 2^{n-1} \]

Step 1: Interpret given sums.
\[ \alpha = \text{sum of even binomial coefficients of } (1+1)^{10} \] \[ \beta = \text{sum of odd binomial coefficients of } (1+1)^{10} \]

Step 2: Use identity.
\[ \alpha = \beta = 2^{9} = 512 \]

Step 3: Find required value.
\[ \alpha - \beta = 512 - 512 = 0 \]