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C4 Binomial Expansion Exam Q

I'm stuck on part (b) of this question:

a) find the binomial expansion of (1+kx)^-2 up to and including the term x^3 where k is a constant

b) the binomial expansion in ascending powers of x of (1+kx)^-2 - (1+x)^n is 6x^2 + px^3 +...
(i) show that n= -2k
(ii) given n<0, find the value of p

Thanks in advance! :smile:

Reply 1

shawn_o1

a) remember the binomial theorem (a + b)^n; apply it to (1 + kx)^n

b) finding k should be easy if you know that the x^2 coefficient in your case is: -2*(-3/2)*(kx)^2 also don't forget about the x^2 coefficient in (1+x)^n
c) apply a)

(edited 9 years ago)

Reply 2

mica-lwe

Original post by shawn_o1

I don't get how that shows that n=-2k ?

Reply 3

shawn_o1

I mean, given that you have 6x^2, find the x^2 coefficients of (1+kx)^-2 and (1+x)^n, then you can find n in terms of k

Reply 4

mica-lwe

Original post by shawn_o1

I mean, given that you have 6x^2, find the x^2 coefficients of (1+kx)^-2 and (1+x)^n, then you can find n in terms of k

I've expanded (1+kx)^-2, so do I expand (1+x)^n or (1+x)^-2k ?

Reply 5

davros

Original post by mica-lwe

I've expanded (1+kx)^-2, so do I expand (1+x)^n or (1+x)^-2k ?

You should be expanding both your original brackets as far as the x^3 term and then comparing the result with the expansion they give you, and using the additional property you proved for part (a).

Reply 6

mica-lwe

Original post by davros

You should be expanding both your original brackets as far as the x^3 term and then comparing the result with the expansion they give you, and using the additional property you proved for part (a).

Sorry I still don't understand :confused:

I've expanded both I just then don't know how to show that n=-2k

Reply 7

shawn_o1

subtract the first expansion with the second expansion, the second expansion should be in terms of n.

Reply 8

mica-lwe

Original post by shawn_o1

subtract the first expansion with the second expansion, the second expansion should be in terms of n.

I've just done that and got

-2k - nx + 3k^2x^2 - (n^2-n(x^2))/2 - 4k^3x^3 - (n^3-3n^2-2n(x^3))/3

but i don't think thats right

Reply 9

shawn_o1

Original post by mica-lwe

I've just done that and got

-2k - nx + 3k^2x^2 - (n^2-n(x^2))/2 - 4k^3x^3 - (n^3-3n^2-2n(x^3))/3

but i don't think thats right

that should actually be 6 (or 3 factorial), after that you're on the right lines

for the terms involving n it may be easier if you don't expand the brackets, that is, leave the terms (n(n-1)/2)x^2 and (n(n-1)(n-2)/6)x^3

(edited 9 years ago)

Reply 10

mica-lwe

Original post by shawn_o1

which part should be 6? and do i make x equal a number so that I just have n and k?

Reply 11

mica-lwe

ignore what I said last, do i use the x^2 coefficients and set them equal to 6?
i did that and got

3K^2 - (n(n-1))/2 = 6

just don't know how to rearrange to n=-2k from there

Reply 12

TenOfThem

Original post by mica-lwe

ignore what I said last, do i use the x^2 coefficients and set them equal to 6?
i did that and got

3K^2 - (n(n-1))/2 = 6

just don't know how to rearrange to n=-2k from there

You do not use that part yet

You use the co-efficient of x to show this

Then you use that and the co-efficient of x^2 to actually find n and k

Then you use that to find p

(edited 9 years ago)

Reply 13

mica-lwe

Original post by TenOfThem

You do not use that part yet

You use the co-efficient of x to show this

Then you use that and the co-efficient of x^2 to actually find n and k

Then you use that to find p