# Binomial Expansions QWatch

#1
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

I have put the x^3 expansions for both together to create:
(-4k^3) - ((n^3)-(3n^2)+(2n))/6 = p <---- I divided by x^3 to get rid of them

If I sub n=-2k then factorise to get solutions of k I get:
k=0 k=-1/4 k=1

If I then sub this into n=-2k I get:
n=0 n=1/2 n=-2

Then I assumed n=-2 because n<0

If I have done that correct, what do I do with n or k?
If I haven't done that correct, what am I supposed to do?

0
4 years ago
#2
(Original post by shuayb96)
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

I have put the x^3 expansions for both together to create:
(-4k^3) - ((n^3)-(3n^2)+(2n))/6 = p <---- I divided by x^3 to get rid of them

If I sub n=-2k then factorise to get solutions of k I get:
k=0 k=-1/4 k=1

If I then sub this into n=-2k I get:
n=0 n=1/2 n=-2

Then I assumed n=-2 because n<0

If I have done that correct, what do I do with n or k?
If I haven't done that correct, what am I supposed to do?

It might be easier if you post a .photo of the question and your workings.
0
4 years ago
#3
(Original post by shuayb96)
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

I have put the x^3 expansions for both together to create:
(-4k^3) - ((n^3)-(3n^2)+(2n))/6 = p <---- I divided by x^3 to get rid of them

If I sub n=-2k
then factorise to get solutions of k I get:
k=0 k=-1/4 k=1

If I then sub this into n=-2k I get:
n=0 n=1/2 n=-2

Then I assumed n=-2 because n<0

If I have done that correct, what do I do with n or k?
If I haven't done that correct, what am I supposed to do?

You can't "sub n =-2k" because you're supposed to be proving that n = -2k!!!

Just write out the first few terms of each binomial expansion and subtract one from the other. Compare the coefficients of what's left with the expansion they have given you.
0
#4
(Original post by davros)
You can't "sub n =-2k" because you're supposed to be proving that n = -2k!!!

Just write out the first few terms of each binomial expansion and subtract one from the other. Compare the coefficients of what's left with the expansion they have given you.
I already proved that n=-2k in part (i)

I was just trying to use that in part(ii) to then calculate the value for 'p'
0
4 years ago
#5
(Original post by shuayb96)
I already proved that n=-2k in part (i)

I was just trying to use that in part(ii) to then calculate the value for 'p'
I can't quite follow what you've done for part(ii). Before you can calculate the value of p, you need to find the correct value of k (and hence n). You haven't used any information about the coefficient of x^2.
0
2 years ago
#6
(Original post by Sniperdon227)
The OP is most likely done with their A-levels now
0
2 years ago
#7
(Original post by SeanFM)
The OP is most likely done with their A-levels now
0
2 years ago
#8
(Original post by Sniperdon227)
Are you stuck with the method?
0
2 years ago
#9
Hi, i am really stuck on the method for both parts (i)&(ii) in b, is anyone abke to come up with a solution?
0
2 years ago
#10
(Original post by RyJParker)
Hi, i am really stuck on the method for both parts (i)&(ii) in b, is anyone abke to come up with a solution?

If you post all your working then you're more likely to get help.
0
2 years ago
#11
(Original post by notnek)

If you post all your working then you're more likely to get help.

https://www.thestudentroom.co.uk/sho....php?t=4454708
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