Question

For a positive integer $n$, if the mean of the binomial coefficients in the expansion of $(a + b)^{2n - 3}$ is 16, then $n$ is equal to :

• A. 5
• B. 7
• C. 9
• D. 4

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the mean of the binomial coefficients and the expansion of the binomial theorem. Let's break down the solution step-by-step:

1. We are given the expression \((a + b)^{2n - 3}\). This expression is expanded using the binomial theorem, which forms binomial coefficients.
2. The binomial coefficient for the term \(k\) is given by: \(\binom{2n-3}{k}\).
3. The number of terms (or the number of binomial coefficients) in the expression \((a + b)^{2n - 3}\) is equal to \(2n - 2\) (since \(k\\) ranges from 0 to \(2n-3\)).
4. The mean of these coefficients is then calculated as: \(\frac{1}{2n-2} \sum_{k=0}^{2n-3} \binom{2n-3}{k}\).
5. According to the binomial theorem, we know that: \(\sum_{k=0}^{2n-3} \binom{2n-3}{k} = 2^{2n-3}\).
6. We are given that this mean is 16: \(\frac{2^{2n-3}}{2n-2} = 16\).
7. This implies that: \(2^{2n-3} = 16 \times (2n-2)\).
8. Since \(16 = 2^4\), we can substitute and simplify: \(2^{2n-3} = 2^4 \times (2n-2)\).
9. Since the powers of 2 on both sides must be equal, we equate: \(2n-3 = 4 + \log_2(2n-2)\).
10. This simplifies to: \(2n - 3 = 4 + \log_2(2n-2)\).
11. Rewriting, we have: \(2n - 7 = \log_2(2n-2)\).
12. By trial and error or calculations, we find that when \(n = 5\), the equation holds: \(2(5) - 7 = \log_2(8)\) which simplifies to \(3 = 3\).

Thus, the correct answer is that \(n\) is equal to 5.