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# P3 revision exercises (and solutions) watch

1. I know beggers can't be choosers, but any chance of the solutions to the Binomial Expansion exercises? In the mean time, could someone tell me if I've got this right:

Expand (1-2x)^-3 up to and including the term in x^3

My answer: 1 + 6x + 24x^2 - 80x^3
2. (Original post by a_r)
I know beggers can't be choosers, but any chance of the solutions to the Binomial Expansion exercises? In the mean time, could someone tell me if I've got this right:

Expand (1-2x)^-3 up to and including the term in x^3

My answer: 1 + 6x + 24x^2 - 80x^3
+80x^3
3. Cheers, thanks. At least I'm heading in the right direction...
4. (Original post by a_r)
I know beggers can't be choosers, but any chance of the solutions to the Binomial Expansion exercises? In the mean time, could someone tell me if I've got this right:

Expand (1-2x)^-3 up to and including the term in x^3

My answer: 1 + 6x + 24x^2 - 80x^3
1+(-3)(-2x)+[(-3)(-4)]/2](-2x)^2+[(-3)(-4)(-5)/6](-2x)^3=

1+6x+24x^2+80x^3
5. My answers to part 1 of Binomial Series NB: Use my answers at your peril; highly likely to contain mistakes

(by the way, just realised I didn't post the link to the Binomial Series exercises earlier in the thread. If anyone does have time to do this exercise, and post their results here, that would be great).

i) 1 + 6x + 24x^2 + 80x^3, mod x<1/2
ii) 1 + 3x - (3/2)x^2 + (5/2)x^3, mod x<1/4
iii) 1 + 4x + 23x^2 + 32x^3, mod x<1/2
iv) 1+ (1/2)x - (1/8)x^2 + (1/16)x^3, mod x<1
v) 1 - 3x + 6x^2 -10x^3, mod x<1
vi) 1 - (1/2)x + (3/4)x^2 - (15/8)x^3, mod x<1
vii) 1 - 2x + 8x^2 - 48x^3, mod x<1/2
viii) 1 - (3/4)x - (3/32)x^2 - (5/128)x^3, mod x<1
ix) 1 + (1/3)x + (5/36)x^2 + (5/81)x^3, mod x<2

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Updated: June 6, 2004
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