Question

The coefficients of `x^2` and `x^3` in the expansion of `(3-2x)^6` are *a* and *b* respectively. Find the values of `a/b` ?

The coefficients of `x^2` and `x^3` in the expansion of `(3-2x)^6` are a and b respectively. Find the values of `a/b` .

Answer 1

The answer is `=1.125`

Explanation:

The binomial theorem is

`(p+q)^6=((6),(0))p^6+((6),(1))p^5q+((6),(2))p^4q^2+((6),(3))p^3q^3+((6),(4))p^2q^4+((6),(5))pq^5+((6),(6))q^6`

where,

`((n),(k))=(n!)/((n-k)!(k!))`

Here,

we have

`p=3` and `q=-2x`

Therefore,

The coefficients of `x^2` is

`=((6),(2))3^4*(-2x)^2=(6!)/((6-2)!(2!))81*(4)x^2`

`=(6*5)/(1*2)*324x^2`

`a=4860`

The coefficients of `x^3` is

`=((6),(3))3^3*(-2x)^3=(6!)/((6-3)!(3!))27*(-8)x^3`

`=-(6*5*4)/(1*2*3)*27*8x^3`

`b=-4320`

Finally,

`a/b=4860/(-4320)=-1.125`