A-Level Maths C4 Binomial expansion question?

Question

Given that in the expansion of (9(1+kx)²)/(3-2x)  the x-coefficient is -1, find the value of x²-coefficient in the form 1/p, where p is an integer. 

That is the question I'm stuck on, I find the terminology somewhat difficult to understand in the context of the question. 
Any help would be greatly appreciated.

fred · Accepted Answer

Hello

9(1+kx)²)/(3-2x)

= 9((1+kx)²)(3-2x)^-1

= 3((1+kx)²)(1 - (2/3)x)^-1   removing factor 3 from final bracket

= 3(1 + 2kx + k²x²)(1 + (-2/3)x)^-1  expanding quadratic

=3(1 + 2kx + k²x²)( 1 + (2/3)x + (4/9)x² + .......)  using the binomial theorem

Looking at the coefficients of the terms in x you have  3 x 1 x 2/3 + 3 x 2k x 1  which is given as -1

so 2 + 6k = -1   k = - ½

Looking at the coefficients of x²  you have 3 x 1 x k²  +  3 x 2k x 2/3  +  3 x 4/9

= 3/4 - 2 + 4/3 = - 1/12  so p = -12

The binomial expansions is valid for  -1 < (-2/3)x < 1  or   -3/2 < x < 3/2

erly · Answer

i'm assuming you may distribute the ^-a million to the two equations in parenthesis first, then distribute the ^0.5 to the hot equations yet i'm no longer rather particular, I discovered a thank you to do Binomials yet i've got under no circumstances completed them with parenthesis like that, i'm in basic terms following the parenthesis rule on that

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A-Level Maths C4 Binomial expansion question?

2 Answers