Q105 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
If the coefficients of $x^2$ and $x^3$ in the expansion of $(3 + kx)^9$ are equal, then the value of '$k$' is
- Correct answerA$\dfrac{9}{7}$
- B$\dfrac{7}{3}$
- C$\dfrac{3}{7}$
- D$\dfrac{7}{9}$
Solution
The general term in the expansion of $(3 + kx)^9$ is $T_{r+1} = ^{9}C_{r} (3)^{9-r} (kx)^r$.
For the coefficient of $x^2$, we put $r = 2$:
Coefficient of $x^2 = ^{9}C_{2} (3)^{7} k^2$
For the coefficient of $x^3$, we put $r = 3$:
Coefficient of $x^3 = ^{9}C_{3} (3)^{6} k^3$
Given that the coefficients are equal:
$^{9}C_{2} (3)^{7} k^2 = ^{9}C_{3} (3)^{6} k^3$
$\dfrac{9 \times 8}{2} \times 3 = \dfrac{9 \times 8 \times 7}{3 \times 2} \times k$
$36 \times 3 = 84 \times k$
$108 = 84k$
$k = \dfrac{108}{84} = \dfrac{9}{7}$
Answer: $\dfrac{9}{7}$
For the coefficient of $x^2$, we put $r = 2$:
Coefficient of $x^2 = ^{9}C_{2} (3)^{7} k^2$
For the coefficient of $x^3$, we put $r = 3$:
Coefficient of $x^3 = ^{9}C_{3} (3)^{6} k^3$
Given that the coefficients are equal:
$^{9}C_{2} (3)^{7} k^2 = ^{9}C_{3} (3)^{6} k^3$
$\dfrac{9 \times 8}{2} \times 3 = \dfrac{9 \times 8 \times 7}{3 \times 2} \times k$
$36 \times 3 = 84 \times k$
$108 = 84k$
$k = \dfrac{108}{84} = \dfrac{9}{7}$
Answer: $\dfrac{9}{7}$
Hence, the correct answer is option: 'A'.