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)

Step 1: Understanding the Concept:
This question tests knowledge of the properties of binomial coefficients, specifically the sums of coefficients with even and odd indices.
Step 2: Key Formula or Approach:
From the binomial expansion of \((1+x)^n\): \[ (1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + \dots + {}^nC_n x^n \] By setting x = 1 and x = -1, we can derive two important identities for `n=10`: 1. Sum of all coefficients: \({}^{10}C_0 + {}^{10}C_1 + \dots + {}^{10}C_{10} = 2^{10}\) 2. Alternating sum: \({}^{10}C_0 - {}^{10}C_1 + {}^{10}C_2 - \dots + {}^{10}C_{10} = (1-1)^{10} = 0\) From the second identity, we can see that the sum of even-indexed coefficients equals the sum of odd-indexed coefficients. Sum of even coefficients: \({}^nC_0 + {}^nC_2 + {}^nC_4 + \dots = 2^{n-1}\) Sum of odd coefficients: \({}^nC_1 + {}^nC_3 + {}^nC_5 + \dots = 2^{n-1}\)
Step 3: Detailed Explanation:
Let's expand the expressions for \(\alpha\) and \(\beta\).
For \(\alpha\), we sum \({}^{10}C_{2k}\) for k = 0, 1, 2, 3, 4, 5: \[ \alpha = {}^{10}C_0 + {}^{10}C_2 + {}^{10}C_4 + {}^{10}C_6 + {}^{10}C_8 + {}^{10}C_{10} \] This is the sum of all binomial coefficients for n=10 with an even lower index. Based on the formula, this sum is equal to \(2^{10-1} = 2^9 = 512\).
For \(\beta\), we sum \({}^{10}C_{2k+1}\) for k = 0, 1, 2, 3, 4: \[ \beta = {}^{10}C_1 + {}^{10}C_3 + {}^{10}C_5 + {}^{10}C_7 + {}^{10}C_9 \] This is the sum of all binomial coefficients for n=10 with an odd lower index. Based on the formula, this sum is also equal to \(2^{10-1} = 2^9 = 512\).
Now, we calculate \(\alpha - \beta\): \[ \alpha - \beta = 512 - 512 = 0 \] Step 4: Final Answer:
The value of \(\alpha - \beta\) is 0.