Question

Suppose A and B are the coefficients of the 30th and 12th terms respectively in the binomial expansion of \( (1 + x)^{2n - 1} \). If \( 2A = 5B \), then \( n \) is equal to:

• A. \( 20 \)
• B. \( 22 \)
• C. \( 21 \)
• D. \( 19 \)

Hint:

When working with binomial expansions, use the general term formula \( T_k = \binom{2n-1}{k} x^k \) to extract the coefficients and find relationships between terms.

Answer 1

The Correct Option is C

In the binomial expansion of \( (1 + x)^{2n - 1} \), the general term is given by: \[ T_k = \binom{2n-1}{k} x^k. \] The 30th term corresponds to \( T_{30} \), and the 12th term corresponds to \( T_{12} \). We are given that \( 2A = 5B \), where \( A \) and \( B \) are the coefficients of the 30th and 12th terms respectively. Solving the equation \( 2A = 5B \), we can find the value of \( n \). 
Final Answer: \( n = 21 \).

Answer 2

Step 1: Write the general term for the binomial expansion.
The general term \( T_{k+1} \) in the expansion of \( (1 + x)^{2n - 1} \) is:
\[ T_{k+1} = \binom{2n - 1}{k} x^k \] where \( k = 0, 1, 2, \dots \)

Step 2: Express the 30th and 12th terms.
- The 30th term corresponds to \( k = 29 \):
\[ A = \binom{2n - 1}{29} \] - The 12th term corresponds to \( k = 11 \):
\[ B = \binom{2n - 1}{11} \]

Step 3: Use the given relation \( 2A = 5B \).
Substitute \( A \) and \( B \):
\[ 2 \binom{2n - 1}{29} = 5 \binom{2n - 1}{11} \] This implies:
\[ \frac{\binom{2n - 1}{29}}{\binom{2n - 1}{11}} = \frac{5}{2} \]

Step 4: Use the formula for binomial coefficients ratio.
\[ \frac{\binom{m}{r}}{\binom{m}{s}} = \frac{m! / (r!(m - r)!)}{m! / (s!(m - s)!)} = \frac{s!(m - s)!}{r!(m - r)!} \] For \( m = 2n - 1 \), \( r = 29 \), \( s = 11 \), this becomes:
\[ \frac{11! (2n - 1 - 11)!}{29! (2n - 1 - 29)!} = \frac{5}{2} \] Simplify factorial expressions accordingly.

Step 5: Simplify and solve for \( n \).
Express the factorial ratio as a product of terms:
\[ \frac{\binom{2n - 1}{29}}{\binom{2n - 1}{11}} = \frac{(2n - 1 - 11)(2n - 1 - 12) \cdots (2n - 1 - 29 + 1)}{(29)(28) \cdots (12)} = \frac{5}{2} \] Counting number of terms in numerator: from \((2n - 12)\) down to \((2n - 29)\) → 18 terms.

After simplification, the equation can be solved for \( n \), and the integer solution is:
\[ \boxed{21} \]