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JEE Mains · Maths · STD 11 - 7. binomial theoram

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

  1. A \(45\)
  2. B \(48\)
  3. C \(51\)
  4. D \(54\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(51\)

Step-by-step Solution

Detailed explanation

Let \(S = \dfrac{2^3}{3} \ ^{12}C_{2} + \dfrac{2^5}{5} \ ^{12}C_{4} + \dfrac{2^7}{7} \ ^{12}C_{6} + \ldots + \dfrac{2^{13}}{13} \ ^{12}C_{12}\) Using the property \(\dfrac{1}{r+1} \ ^{n}C_{r} = \dfrac{1}{n+1} \ ^{n+1}C_{r+1}\), we can rewrite the terms:…
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