The cofficient of x^(-7) in the expansion of (1+x)^12 (1+1/x)^12 is

Question

this is the qustion is the binomal expansion

Answer 1

(1+x)(1+1/x)=(1+1/x+x+1)=(2+1/x+x)=(x^2+2x+1)/x=(x+1)^2/x.

Therefore (1+x)^12(1+1/x)^12=(1+x)^24/x^12.

Everything in the expansion of the numerator is divided by x^12 so if x^a is the desired term, x^a/x^12=x^-7. Therefore, a-12=-7, and a=5. So we need the coefficient of x^5.

The expansion starts 1+24x+276x^2+2024x^3+10626x^4+42504x^5+...

Therefore the coefficient of x^-7 is 42504.

(The coefficients are calculated thus: 24, 24*23/2!=276, 24*23*22/3!=24*23*22/(3*2)=2024, and so on.)