We have, (3+x+x2)6 General term of the expansion =
6!
r!s!t!
×3r×x5×(x2)t Where, r+s+t=6 =
6!
r!s!t!
×3r×x5+2t For the coefficient of x5, s+2t=5 But r+s+t=6 r+5−t=6 r−a=1+t and s=5−2t Now, t=0, then r=1,s=5 t=1, then r=2,s=3 t=2 then r=3,s=1 ∴ There are only 3 terms that contain x5 ∴ Coefficient of x5 =
