The coefficients of $x^{2}$ $x^2$ and $x^{3}$ $x^3$ in the expansion of ${(3 - 2 x)}^{6}$ $(3-2x)^6$ are a and b respectively. Find the values of $\frac{a}{b}$ $a/b$ ?

Question

The coefficients of $x^{2}$ $x^2$ and $x^{3}$ $x^3$ in the expansion of ${(3 - 2 x)}^{6}$ $(3-2x)^6$ are a and b respectively. Find the values of $\frac{a}{b}$ $a/b$ .

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Answer 1 · 2017-11-25T15:55:23

The answer is $= 1.125$ $=1.125$

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,

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