What will you be doing over the Easter break? >>

C4 Binomial expansion help please! :)

Bit stuck on how to to this..please help! :smile:

So...

When (1+ax)^n is expanded as a series in ascending powers of x, the coefficient of x and x^2 are -6 and 27 respectively.
Find the values of a and n

Thanks :smile:

Reply 1

jase951

use the formula for binomial expansion. but this time where there is X in the formula u need to have aX.

Reply 2

Ballet_shoes

Yh...I did that and got na=-6 and [n(n-1)/2!]a^2=27
Then what happens or do I not need to do that?

Reply 3

amar247

yeh dats right mate. change na=6 to a=6/n and sub into second equation, u get a quadratic in n, work out n and coresspondin a value. if there are no restrictions on n or a , then both sets of value go. but normally there is sum sore of restriction

Reply 4

amar247

actually sorry read the question, u change na=-6 to a=-6/n, sub into [n(n-1)/2!]a^2 and work out n, u do not get a quadractic

Reply 5

Ballet_shoes

thanks thanks! Got it! :smile:

Reply 6

amar247

n = -8 and a = -3/4, thought id let you know

Reply 7

Ballet_shoes

Answers are at the back of the book, it says n=-2 and a=3
I have n=-1.5 and a=4
Haha
Wonder who's right? This is soo fustratingg!! Lol

Reply 8

Azmosis

compare the co-efficients of both x and x^2, not just x^2.

Reply 9

Ballet_shoes

Ok worked out that the answer in the back of the textbook is right...I still can't get it though...lol
*sigh*

Reply 10

annikapaula

1 + na x + n(n-1)a x^2 = 1 - 6x + 27 x^2

compare coefficient

na x = -6 x
[n(n-1)a x^2] / 2! = 27 x^2

||
\/

na = -6
[n(n-1)a^2] / 2! = 27

n = -36 / 18 = -2
a = -6 / -2 = 3

:biggrin:

Reply 11

TeeEm

Original post by annikapaula

1 + na x + n(n-1)a x^2 = 1 - 6x + 27 x^2

compare coefficient

na x = -6 x
[n(n-1)a x^2] / 2! = 27 x^2

||
\/

na = -6
[n(n-1)a^2] / 2! = 27

n = -36 / 18 = -2
a = -6 / -2 = 3

:biggrin:

do you realize you are answering a question from 2009?

Reply 12

Jordan97

I've seen questions like this in C2 papers

Reply 13

annikapaula

Original post by TeeEm

do you realize you are answering a question from 2009?

YUPPY!

It might not be helpful for the one who asked but it might be for those who will encounter this! :tongue:

Like me, I am currently in A2 and my teacher gave us this question in one of our worksheets ^_^ I search in the internet but I can't find this question in any past papers and only this site has the answers but no explanations :smile:

You need to figure it out by yourself! Still even the answers are given, some students will still not know how the answers is calculated :smile:

Reply 14

Horace970

36(n^2-n) / 2n^2 =27

36n^2-36n=54n^2

n^2=2n=0
n=-2
then sub 'n' back into an=-6
therefore, -2a=-6
thus a=3.

Reply 15

Kernowtom

Original post by annikapaula

1 + na x + n(n-1)a x^2 = 1 - 6x + 27 x^2

compare coefficient

na x = -6 x
[n(n-1)a x^2] / 2! = 27 x^2

||
\/

na = -6
[n(n-1)a^2] / 2! = 27

n = -36 / 18 = -2
a = -6 / -2 = 3

:biggrin:

Why did you divide by 18 and -2?