HKCEE 1992

In the expansion of (1+3x)^2(1+x)^n,where n is a pssitive integer, the coefficient of x is 10.
a) Find the value of n.
b) Find the coefficient of x^2.
(a)
(1+3x)^2(1+x)^n
=(1+6x+9x^2)(1+nx+(n(n-1)/2)x^2+...)
=1+(n+6)x+[n(n-1)/2+6n+9]x^2+...
the coefficient of x is 10.
n+6=10
n=4
b)
the coefficient of x^2
=n(n-1)/2+6n+9
=4*3/2+4*6+9
=6+24+9
=39

(1+3x)^2(1+x)^n
=(1+6x+9x^2)*(1+nx+n*(n-1)/2 * x^2...)
=1+6x+ nx+9x^2 +6nx^2+n*(n-1)/2*x^2
a) 6+n=10
n=4
b)the coefficient of x^2 is
9 +6n+n*(n-1)/2
=9+6*4 + 4*3/2
=9+24+6
=39

Solution

(a)
(1+3x)^2(1+x)^n
= (1 + 6x + 9x^2) [1 + (nC1)x + (nC2)x^2 + ......]

Coefficient of x
= 1 (nC1) + 1 (6)
= 1 (n) + 6
= n + 6

So,
n + 6 = 10
n = 4

(b)
When n = 4
Coefficient of x^2
= 1 (4C2) + 6 (4C1) + 9 (1)
= 6 + 6 (4) + 9
= 39