Question

The coefficients of x^2 and x^3 in the expansion of (3 − 2x) ^6 are a and b respectively. What is the value of a/B?

Answer 1

By binomial theorem

$(x+y)^n=\sum\limits_{k=0}^{n}\begin{pmatrix} n \\ k \end{pmatrix}x^{n-k}y^k=\\ \sum\limits_{k=0}^{n}\cfrac{n!}{(n-k)!k!}x^{n-k}y^k\\ ax^2=\cfrac{6!}{(6-2)!2!}3^{4}(-2x)^2=4860x^2\\ bx^3=\cfrac{6!}{(6-3)!3!}3^{3}(-2x)^3=-4320x^3\\ \cfrac{a}{b}=\cfrac{4860}{-4320}=-1.125$