Binomial Theorem

Question

In the expansion of (2+x)^14 multiplied by (1+2/x)^14, find the coefficient of x^12.

Answer 1

Coefficients of first expansion:

2^14 + (14C1)*2^13*x + (14C2)*2^12*x^2 + ... (14C12)*2^2*x^12 + (14C13)*2*x^13 + x^14             (1)

Coefficients of the second expansion:

1 + (14C1)*1*(2/x) + (14C2)*2^2*(1/x)^2 + ... + 2^14*(1/x)^14         (2)

 

When multiplying these expanded values, the x^12 term is when the x^12 of the first is multiplied by the constant of the second, also, when the x^13 term of (1) is multiplied by the 1/x term of the second, and when the x^14 of the first is multiplied by the (1/x)^2 term, thus:

 

[(14C12)*2^2*x^12] x [1] + [(14C13)*2*x^13] x [(14C1)*(2/x)] + [x^14] x [(14C2)*2^2*(1/x)^2]

=> 14C12 is (14*13)/2 = 91; 14C13 = 14; 14C1 = 14; 14C2 = 91;

 

[91*4 + 14*2*14*2 + 91*4]*x^12

 

=> 1512*x^12

So the coefficient is 1512.