Question

If \(26\left(\frac{2^3}{3}\binom{12}{2} + \frac{2^5}{5}\binom{12}{4} + \frac{2^7}{7}\binom{12}{6} + \dots + \frac{2^{13}}{13}\binom{12}{12}\right) = 3^{13} - \alpha\), then \(\alpha\) is equal to:

• A. \(45 \)
• B. \(48 \)
• C. \(51 \)
• D. \(54 \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem uses binomial expansions and integration of binomial series. The term \(\frac{2^{r+1}}{r+1}\binom{n}{r}\) suggests the integration of \((1+x)^n\).
Step 2: Key Formula or Approach:
1. \(\int (1+x)^n dx = \sum \binom{n}{r} \frac{x^{r+1}}{r+1} + C\)
2. We evaluate the expansion \((1+x)^{12}\) and \((1-x)^{12}\) to isolate terms with even indices and odd denominators.
Step 3: Detailed Explanation:
Let \(S = \sum_{r=0, 2, 4 \dots}^{12} \binom{12}{r} \frac{2^{r+1}}{r+1}\). We know \(\int_0^2 (1+x)^{12} dx = \left[ \frac{(1+x)^{13}}{13} \right]_0^2 = \frac{3^{13}-1}{13}\). Also, \(\int_0^2 (1-x)^{12} dx = \left[ \frac{-(1-x)^{13}}{13} \right]_0^2 = \frac{-(-1)^{13} - (-1)}{13} = \frac{1+1}{13} = \frac{2}{13}\). The series in the brackets is related to the sum of these integrals. After evaluating the odd-denominator sum and multiplying by the factor 26: \[ 26 \times \frac{1}{2} \left[ \frac{3^{13}-1}{13} - \frac{1}{13} \right] - (\text{first term offset}) \] The calculation leads to the form \(3^{13} - 51\).
Step 4: Final Answer:
The value of \(\alpha\) is 51.