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When ( 1 - x)(1 + ax)^6 is expanded as far as the term in x^2, the result is 1 + bx^2. Find the value of a and b.
My workings are as follows : (ain't sure whether I'm correct)
(1 - x)(1 + 6ax +15a^2x^2)
Use this to get terms of up to powers of 2.
(1 - 6ax^2 + 15a^2x^2) = 1 + bx^2
I'm stuck after this. I only got as far as (15a^2 - 6a) =b
answer's a=1/6 and b = - 7/12.
Please help. Thanks so much.
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Originally posted by secretliker:
(1-x)(1+ax)^6
(1-x)(1+6ax+15a^2x^2)
1 + 6ax + 15a^2x^2 - x - 6ax^2 - 15a^2x^3
1 + (6a-1)x - 6ax^2 + 15a^2x^2 = RHS
1 + (6a-1)x - 6ax^2 + 15a^2x^2 = 1 + bx^2Now it can be shown that 6a-1 = 0
Thus a = 1/6
15a^2 - 6a = b
b = -7/12
Oh, now i got it! So, does this mean my working for
(1 - 6ax^2 + 15a^2x^2) = 1 + bx^2 is wrong and redundant. cuz i didn't get the x^1 out and hence wasn't able to deduce the value of a?
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Originally posted by bonkysleuth:
Oh, now i got it! So, does this mean my working for
(1 - 6ax^2 + 15a^2x^2) = 1 + bx^2 is wrong and redundant. cuz i didn't get the x^1 out and hence wasn't able to deduce the value of a?
You expanded wrongly.
The question is
expanded as far as the term in x^2
You didn't expand the term in x, which would have given you the value of a. So yup, your final statement is correct.
Edited by eagle 08 Jun `08, 10:44AM
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