Question

If the coefficients of \( x^3 \) and \( x^4 \) in the expansion of \( (3+kx)^9 \) are equal, then the value of \( k \) is:

• A. \( 3 \)
• B. \( \frac{1}{3} \)
• C. \( 2 \)
• D. \( \frac{1}{2} \)
• E. \( 1 \)

Hint:

In the expansion of \( (a+bx)^n \), if the coefficients of \( x^r \) and \( x^{r+1} \) are equal, then \( k = \frac{a(r+1)}{b(n-r)} \). Using this shorthand saves massive amounts of calculation.

Answer 1

The Correct Option is C

Concept: The general term in the binomial expansion of \( (a+b)^n \) is \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \). Here, we compare the terms involving \( x^3 \) and \( x^4 \).

Step 1: Find the coefficient of \( x^3 \).
For \( x^3 \), we set \( r = 3 \) in the expansion of \( (3+kx)^9 \): \[ T_4 = \binom{9}{3} (3)^{9-3} (kx)^3 = \binom{9}{3} 3^6 k^3 x^3 \] Coefficient of \( x^3 = \binom{9}{3} 3^6 k^3 \).

Step 2: Find the coefficient of \( x^4 \).
For \( x^4 \), we set \( r = 4 \): \[ T_5 = \binom{9}{4} (3)^{9-4} (kx)^4 = \binom{9}{4} 3^5 k^4 x^4 \] Coefficient of \( x^4 = \binom{9}{4} 3^5 k^4 \).

Step 3: Equate the two coefficients and solve for \( k \).
\[ \binom{9}{3} 3^6 k^3 = \binom{9}{4} 3^5 k^4 \] Divide both sides by \( 3^5 k^3 \): \[ \binom{9}{3} \times 3 = \binom{9}{4} \times k \] Using the formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \): \[ \frac{9!}{3!6!} \times 3 = \frac{9!}{4!5!} \times k \] \[ \frac{3}{6 \times 1} = \frac{k}{4 \times 6} \quad \text{(simplified after canceling 9! and common factors)} \] Alternatively, using \( \binom{n}{r} = \frac{n-r+1}{r} \binom{n}{r-1} \): \[ k = 3 \times \frac{\binom{9}{3}}{\binom{9}{4}} = 3 \times \frac{4}{9-4+1} = 3 \times \frac{4}{6} \] \[ k = 3 \times \frac{2}{3} = 2 \]