First three terms in expansion of (1 − 2x)2(1 + ax)6, in ascending powers of x, are 1 − x + bx2. Find a and b?

Answers

Answer #1 | 07/02 2014 06:13

a = 1/2; b = - 17/4
f(x) = (1 − 2x)^2(1 + ax)^6
f '(x) = 2( -2) (1 - 2x) (1 + ax)^6 + (1 − 2x)^2 6a(1 + ax)^5 =
= 2(1 - 2x)(1 + ax)^5 ( -2(1 + ax) + 3a(1 - 2x) ) =
f '(x) = 2 (2 x - 1) (8 a x - 3 a + 2) (a x + 1)^5
f ''(x) =
2 (112 a^2 x^2 + 28 a x (2 - 3 a) + 15 a^2 - 24 a + 4) (a x + 1)^4
f(0) = 1
f '(0) = 2 (3 a - 2)
f ''(0) = 2 (15 a^2 - 24 a + 4)
Taylor expansion is
1 + 2x(3a - 2) + x^2 (15a^2 - 24a + 4)
coefficient of x is 2(3a - 2) = - 1
a = 1/2
plug in the coefficient of x^2
15(1/2)^2 - 24(1/2) + 4 = - 17/4

Positive: 45 %

Answer #2 | 07/02 2014 05:55

First three terms in expansion of (1 - 2x)^2(1 + ax)^6,
in ascending powers of x, are 1 - x + bx2.
1 - 4x + 4x^2
b = 4
Find a and b

Positive: 39 %

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Answer #1| 07/02 2014 06:1345 %