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C2 bionomial expansion

I am stuck on this question:
given that (1+x)^n = 1 + 12x + x^2 find the value of a

Does n= 2 ? Im not sure on how to do this question
Reply 1
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Robbie242
Original post by Alex-Torres
I am stuck on this question:
given that (1+x)^n = 1 + 12x + x^2 find the value of a

Does n= 2 ? Im not sure on how to do this question
Where's a being defined? all I can see are n's and x's, unless you mean (1+ax) but still then its a bit cryptic. And no, n does not equal 2 since using the laws of expansion
(1+x)n=1n+nx+....(1+x)^n=1^n+nx+....
(edited 11 years ago)
Original post by Alex-Torres
I am stuck on this question:
given that (1+x)^n = 1 + 12x + x^2 find the value of a

Does n= 2 ? Im not sure on how to do this question

There isn't an 'a' in the question?
Reply 3
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the bear
20
was it asking for

( 1+ax )n ?
Reply 4
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Alex-Torres
OP
15
Original post by Robbie242
Where's a being defined? all I can see are n's and x's, unless you mean (1+ax) but still then its a bit cryptic. And no, n does not equal 2 since using the laws of expansion
(1+x)n=1n+nx+....(1+x)^n=1^n+nx+....


Original post by Felix Felicis
There isn't an 'a' in the question?


Original post by the bear
was it asking for

( 1+ax )n ?

Sorry. Given that (1+x)^n = 1 + 12x + ax^2, find the value of a.
Reply 5
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Robbie242
Original post by Alex-Torres
Sorry. Given that (1+x)^n = 1 + 12x + ax^2, find the value of a.
Notice that from earlier the 2nd term 12x=nx12x=nx divide by x, and we have n=12, I'm sure you know how to solve for a now

Spoiler

(edited 11 years ago)
Reply 6
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the bear
20
well the x2 term is given by

{n( n-1 )/2! }x2
Original post by Alex-Torres
Sorry. Given that (1+x)^n = 1 + 12x + ax^2, find the value of a.

Should this be (1+x)n=1+12x+ax2+...(1+x)^{n} = 1 + 12x + ax^{2} + ... or does it stop at the x^2 term?
Reply 8
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Alex-Torres
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15
Original post by Felix Felicis
Should this be (1+x)n=1+12x+ax2+...(1+x)^{n} = 1 + 12x + ax^{2} + ... or does it stop at the x^2 term?

It stops at the ax^2
Reply 9
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Robbie242
Original post by Alex-Torres
It stops at the ax^2
That can't be right? By the information you've given me I've deduced that n=12, it can't be untrue as its part of the standard expansion, I think it goes on for longer otherwise people would just say n=2 herp derp which isn't true for the 2nd term
(edited 11 years ago)
Reply 10
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Alex-Torres
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15
Original post by Robbie242
Notice that from earlier the 2nd term 12x=nx12x=nx divide by x, and we have n=12, I'm sure you know how to solve for a now

Spoiler


I got 66?
image.jpg145.6KB
Reply 11
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Robbie242
Original post by Alex-Torres
I got 66?
That is correct, now set it equal to ax^2 and solve for a also a quicker way for terms like this n(n−1)2!(x2)\frac{n(n-1)}{2!}(x^2) if in the form (1+x)n(1+x)^n
(edited 11 years ago)
Reply 12
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Alex-Torres
OP
15
Original post by Robbie242
That is correct, now set it equal to ax^2 and solve for a also a quicker way for terms like this n(n−1)2!(x2)\frac{n(n-1)}{2!}(x^2) if in the form (1+x)n(1+x)^n

Doesnt a just equal 66 and thats the end of the question?
Reply 13
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Robbie242
Original post by Alex-Torres
Doesnt a just equal 66 and thats the end of the question?
and that is indeed, try not to forget if you expand up to a term to leave +.... some examiners may be very picky about this
Reply 14
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Alex-Torres
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15
Original post by Robbie242
and that is indeed, try not to forget if you expand up to a term to leave +.... some examiners may be very picky about this

Thanks!
Do you know where I went wrong in this question?image.jpg
Reply 15
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Robbie242
Original post by Alex-Torres
Thanks!
Do you know where I went wrong in this question?image.jpg
Try be more careful in your workings and re-expand, one of the terms I got was to be -45, just re-expand more carefully
Reply 16
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Alex-Torres
OP
15
Original post by Robbie242
Try be more careful in your workings and re-expand, one of the terms I got was to be -45, just re-expand more carefully

Could you please take a picture of your working, as I can't see how you got that answer?
Reply 17
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Robbie242
Original post by Alex-Torres
Could you please take a picture of your working, as I can't see how you got that answer?
Sure after dinner if your still stuck, will be like 10 mins
Original post by Alex-Torres
Thanks!
Do you know where I went wrong in this question?image.jpg

Man, root3^5 is root3 x root3 x root3 x root3 x root 3... Which gives root 243 which is 9root 3. :tongue:
And similarly for the rest! :biggrin:
Reply 19
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Alex-Torres
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15
Original post by Robbie242
Sure after dinner if your still stuck, will be like 10 mins

Im OK now, but thanks anyway! And the above poster

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