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:
The presence of binomial coefficients divided by odd numbers (\( 3, 5, 7 \dots \)) indicates that the sequence originates from the integration of a binomial expansion. The alternating combination suggests we need to add two binomial series before integrating.
Step 2: Key Formula or Approach:
Use the standard expansion identity for even powers:
\( (1+x)^n + (1-x)^n = 2 \left[ \binom{n}{0} + \binom{n}{2}x^2 + \binom{n}{4}x^4 + \dots \right] \)
Integrate both sides with respect to \( x \) from \( 0 \) to the required upper limit to match the powers in the question.
Step 3: Detailed Explanation:
Let the given sum inside the parentheses be \( S \):
\( S = \frac{2^3}{3} \binom{12}{2} + \frac{2^5}{5} \binom{12}{4} + \dots + \frac{2^{13}}{13} \binom{12}{12} \).
Set \( n = 12 \) in our identity:
\( (1+x)^{12} + (1-x)^{12} = 2 \left[ \binom{12}{0} + \binom{12}{2}x^2 + \binom{12}{4}x^4 + \dots + \binom{12}{12}x^{12} \right] \).
Integrate both sides with respect to \( x \) from \( 0 \) to \( 2 \):
\( \int_0^2 \left( (1+x)^{12} + (1-x)^{12} \right) dx = 2 \int_0^2 \left[ \binom{12}{0} + \binom{12}{2}x^2 + \dots + \binom{12}{12}x^{12} \right] dx \).
Evaluate the Left Hand Side (LHS):
\( \text{LHS} = \left[ \frac{(1+x)^{13}}{13} - \frac{(1-x)^{13}}{13} \right]_0^2 \)
\( = \left( \frac{3^{13}}{13} - \frac{(-1)^{13}}{13} \right) - \left( \frac{1^{13}}{13} - \frac{1^{13}}{13} \right) \)
\( = \frac{3^{13} + 1}{13} - 0 = \frac{3^{13} + 1}{13} \).
Now evaluate the Right Hand Side (RHS):
\( \text{RHS} = 2 \left[ \binom{12}{0}x + \binom{12}{2}\frac{x^3}{3} + \binom{12}{4}\frac{x^5}{5} + \dots + \binom{12}{12}\frac{x^{13}}{13} \right]_0^2 \)
\( = 2 \left[ \binom{12}{0}(2) + \binom{12}{2}\frac{2^3}{3} + \binom{12}{4}\frac{2^5}{5} + \dots + \binom{12}{12}\frac{2^{13}}{13} \right] \).
Notice that the terms from the second one onwards exactly form our sum \( S \).
\( \text{RHS} = 2 \left[ 2(1) + S \right] = 4 + 2S \).
Equating LHS and RHS:
\( \frac{3^{13} + 1}{13} = 4 + 2S \)
Multiply by 13:
\( 3^{13} + 1 = 52 + 26S \)
\( 26S = 3^{13} + 1 - 52 \implies 26S = 3^{13} - 51 \).
The original equation is given as \( 26S = 3^{13} - \alpha \).
Comparing both expressions, we find \( \alpha = 51 \).
Step 4: Final Answer:
The value of \( \alpha \) is 51.