If in the expansion of ${(1 + a x)}^{n}$ $(1+ax)^n$ coefficient of $x^{3}$ $x^3$ is six times that of $x^{2}$ $x^2$ , find $a$ $a$ and $n$ $n$ ?

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Answer 1 · 2017-02-04T14:43:30

$a = - 3$ $a=-3$ and $n = - 4$ $n=-4$

Binomial expansion of ${(1 + a x)}^{n}$ $(1+ax)^n$ is

$1+^nC_1ax+^nC_2(ax)^2+^nC_3(ax)^3+........+^nC_n(ax)^n$ ,

where $color(white)(x)^nC_r=(n!)/(r!(n-r)!)$ or $(n(n-1)(n-2)....(n-r+1))/(1xx2xx3xx...xxr)$ and $a$ is a constant.

As coefficient of $x^3$ is six times the coefficient of $x^2$

$color(white)(x)^nC_3xxa^3=6xx^nC_2xxa^2$

i.e. $(n(n-1)(n-2))/(1xx2xx3)xxa=6xx(n(n-1))/(1xx2)$

or $(n-2)/3xxa=6$ i.e. $a(n-2)=18$ .................(1)

Further, coefficient of $x$ is $12$ , we have

$color(white)(x)^nC_1xxa=12$ i.e. $axxn=12$ .................(2)

Putting (2) in (1) we get $12-2a=18$

Hence, $a=(18-12)/(-2)=-3$

and $n=12/-3=-4$

and hence expansion is $(1-3x)^(-4)$

= $1+(-4)/1(-3x)+((-4)(-5))/2(-3x)^2+((-4)(-5)(-6))/6(-3x)^3+................$

= $1+12x+90x^2+540x^3+..........$

If in the expansion of (1+ax)^n(1+ax)n(1+ax)^n coefficient of x^3x3x^3 is six times that of x^2x2x^2, find aaa and nnn?