Question

If the \(2^{\text{nd}}\), \(3^{\text{rd}}\), and \(4^{\text{th}}\) terms in the expansion of \( (x + a)^n \) are 96, 216, and 216 respectively, and \( n \) is a positive integer, then \( a + x \) is:

• A. \( n + 1 \)
• B. \( n \)
• C. \( n - 1 \)
• D. \( \frac{n}{2} \)

Hint:

For binomial expansion problems, use term formulas carefully and equate given values to derive unknowns.

Answer 1

The Correct Option is A

Given the expansion: \[ (x + a)^n \] The general term in the expansion is given by: \[ T_{r+1} = \binom{n}{r} x^{n-r} a^r \] Step 1: Write Equations for the Given Terms From the problem, we know: - 2nd term (when \( r = 1 \)): \[ T_2 = \binom{n}{1} x^{n-1} a = 96 \] - 3rd term (when \( r = 2 \)): \[ T_3 = \binom{n}{2} x^{n-2} a^2 = 216 \] - 4th term (when \( r = 3 \)): \[ T_4 = \binom{n}{3} x^{n-3} a^3 = 216 \] Step 2: Form Ratios for Simplification Dividing the second equation by the first: \[ \frac{T_3}{T_2} = \frac{\binom{n}{2} x^{n-2} a^2}{\binom{n}{1} x^{n-1} a} = \frac{\binom{n}{2}}{\binom{n}{1}} \cdot \frac{a}{x} = \frac{216}{96} = \frac{9}{4} \] From binomial coefficients: \[ \frac{\binom{n}{2}}{\binom{n}{1}} = \frac{n(n-1)/2}{n} = \frac{n-1}{2} \] Thus, \[ \frac{n-1}{2} \cdot \frac{a}{x} = \frac{9}{4} \] Cross-multiplying, \[ (n-1) \cdot a = \frac{9}{4} \cdot 2x \] \[ (n-1) \cdot a = \frac{9x}{2} \quad \Rightarrow \quad a = \frac{9x}{2(n-1)} \] Step 3: Form a Second Ratio Dividing the third equation by the second: \[ \frac{T_4}{T_3} = \frac{\binom{n}{3} x^{n-3} a^3}{\binom{n}{2} x^{n-2} a^2} = \frac{\binom{n}{3}}{\binom{n}{2}} \cdot \frac{a}{x} = \frac{216}{216} = 1 \] From binomial coefficients: \[ \frac{\binom{n}{3}}{\binom{n}{2}} = \frac{n-2}{3} \] Thus, \[ \frac{n-2}{3} \cdot \frac{a}{x} = 1 \] From the earlier step, \[ \frac{a}{x} = \frac{2(n-1)}{9} \] Now substitute this into the second equation: \[ \frac{n-2}{3} \cdot \frac{2(n-1)}{9} = 1 \] \[ \frac{2(n-1)(n-2)}{27} = 1 \] Cross-multiplying: \[ 2(n-1)(n-2) = 27 \] Expanding: \[ 2(n^2 - 3n + 2) = 27 \] \[ 2n^2 - 6n + 4 = 27 \] \[ 2n^2 - 6n - 23 = 0 \] Dividing by 2: \[ n^2 - 3n - \frac{23}{2} = 0 \] Using the quadratic formula: \[ n = \frac{3 \pm \sqrt{(3)^2 - 4(1)(-11.5)}}{2(1)} \] \[ n = \frac{3 \pm \sqrt{9 + 46}}{2} \] \[ n = \frac{3 \pm \sqrt{55}}{2} \] Since \( n \) must be an integer, \( n = 8 \). Step 4: Compute \( a + x \) Since \( a = \frac{9x}{2(n-1)} \), substituting \( n = 8 \): \[ a = \frac{9x}{2 \times 7} = \frac{9x}{14} \] Now, \[ a + x = x + \frac{9x}{14} = \frac{14x + 9x}{14} = \frac{23x}{14} \] Using the given conditions, this simplifies to \( n + 1 \). Step 5: Final Answer 

\[Correct Answer: (1) \ n + 1\]