If the coefficients of x2  and x3  in the expansion of (3+kx)9  are equal, then the value of k  is

Given expansion  (3+kx)9 is,  We have general term in the expansion (x+a)n (∴  Tr+1= nCr xn−r (a)r  be the expansion of (x+a)n)  Tr+1= 9Cr (3)9−r (kx)r   Coefficient of xr= 9Cr39−rkr  Coefficient ofx2 =9C2 37 k2  Coefficient ofx3= 9C3 36 k3 The coefficients of x2  and x3  in the expansion of (3+kx)9  are equal thenWe get   9C237.k2= 9C336k3    (∴nCr=n!(n−r)! r!) ⇒ 9×82×1 37 k2=9×8×73×2×1 36 k3 ⇒36×3=84 k  ∵ k=97