Question

When \( |x|>3 \), the coefficient of \( \frac{1}{x^n} \) in the expansion of \( x^{3/2} (3+x)^{1/2} \) is

• A. \( (-1)^n \frac{1.3.5 \dots (2n-1)}{2^n n!} 3^n \)
• B. \( (-1)^{n+1} \frac{1.3.5 \dots (2n+1)}{2^{n+2} (n+2)!} 3^{n+2} \)
• C. \( (-1)^{n+1} \frac{1.3.5 \dots (2n-1)}{2^n n!} 3^{n+1} \)
• D. \( (-1)^{n+1} \frac{1.3.5 \dots (2n+1)}{2^{n+3} (n+2)!} 3^{n+1} \)

Hint:

For binomial expansions of \( (a+x)^n \) where \( n \) is negative or fractional, always factor out the term with the larger magnitude to ensure the expansion converges (ratio \(<1 \)).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

To expand \( (3+x)^{1/2} \) using the binomial theorem for rational indices, we must ensure the term in the expansion is less than 1. Given \( |x|>3 \), we have \( |\frac{3}{x}|<1 \). Therefore, we should factor out \( x \) from the binomial term.
Step 2: Detailed Explanation:

Rewrite the expression: \[ E = x^{3/2} (x + 3)^{1/2} = x^{3/2} \left[ x \left( 1 + \frac{3}{x} \right) \right]^{1/2} \] \[ E = x^{3/2} \cdot x^{1/2} \left( 1 + \frac{3}{x} \right)^{1/2} = x^2 \left( 1 + \frac{3}{x} \right)^{1/2} \] We need the coefficient of \( \frac{1}{x^n} \), i.e., \( x^{-n} \). Let the general term in the binomial expansion of \( (1 + y)^m \) be \( T_{r+1} \). \[ T_{r+1} = \frac{m(m-1)\dots(m-r+1)}{r!} y^r \] Here \( m = 1/2 \) and \( y = 3/x \). The term in the full expression is: \[ x^2 \cdot T_{r+1} = x^2 \cdot \binom{1/2}{r} \left( \frac{3}{x} \right)^r = \binom{1/2}{r} 3^r x^{2-r} \] We want the power of \( x \) to be \( -n \): \[ 2 - r = -n \implies r = n + 2 \] Now, calculate the coefficient \( \binom{1/2}{r} \) for \( r = n+2 \): \[ \binom{1/2}{r} = \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)\dots(\frac{1}{2}-r+1)}{r!} \] \[ = \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})\dots(-\frac{2r-3}{2})}{r!} \] The numerator has \( r \) terms. The first term is positive, and the remaining \( r-1 \) terms are negative. So the sign is \( (-1)^{r-1} \). \[ \binom{1/2}{r} = \frac{(-1)^{r-1} \cdot 1 \cdot 1 \cdot 3 \cdot 5 \dots (2r-3)}{2^r r!} \] Substituting \( r = n+2 \): \[ \text{Coeff} = \left[ \frac{(-1)^{n+1} \cdot 1 \cdot 3 \cdot 5 \dots (2(n+2)-3)}{2^{n+2} (n+2)!} \right] \cdot 3^{n+2} \] The last factor in the numerator product is \( 2n+4-3 = 2n+1 \). \[ \text{Coeff} = (-1)^{n+1} \frac{1 \cdot 3 \cdot 5 \dots (2n+1)}{2^{n+2} (n+2)!} 3^{n+2} \]
Step 4: Final Answer:

The correct option matches this expression.