Question

if the coefficient of x² and x³ in the expansion of (3+kx)^9 are equal find k​

Answer 1

Answer:

the (r+h)th term in the expansion of (a+b)n is

TR+1 =NCR an-r be

(r+h)th term in the expansion of (3+kx)9 is

TR+1 = 9cr (3)9-r (kx)r

=9cr (3)9-r (kx)

in this is the term containg x² then r=2 and coefficient of x² = 9c(3)7 k²

since the coefficient of x² and x³ are equal we have form +(2) and (3)

9c2 (3)7k²=9c3 (3)^6 k³

9!/2! 7!×3^7 /3^6 = 9!/3! 6! × k³/k²

1/2! ×7× 6! ×3 =1/3×2!×6 ×k

3/7=k/3

$k \frac{ 9}{7}$