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# Binomial expansion question Watch

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1. Can anyone answer this and explain how you got there as I'm struggling to even start answering this question.

2. (Original post by Alb1)
Can anyone answer this and explain how you got there as I'm struggling to even start answering this question.

The general term for (x^4 - 1/x^2)^5 is 5Cr * (x^4)^(5-r) * (-1/x^2)^r = 5Cr * x^(20-4r) * (-1)^r * x^(-2r) = 5Cr * (-1)^r * x^(20-6r), for r = 0, 1, 2, 3, 4, 5.
Thus the powers of x will be 20, 14, 8, 2, -4, and -10.
Similarly the general term for (x - 1/x)^6 is 6Cr * x^(6-r) * (-1/x)^r = 6Cr * (-1)^r * x^(6-2r), for r = 0, 1, 2, 3, 4, 5, 6.
Thus the powers of x will be 6, 4, 2, 0 (constant), -2, -4, and -6.
Now notice that x^(-12) must therefore come from x^(-10) * x^(-2). x^(-10) comes from r=5 in the first expansion, so its coefficient is 5C5 * (-1)^5, which is 1 * -1 = -1. x^(-2) comes from r=4 in the second expansion, so its coefficient is 6C4 * (-1)^4 = 15 * 1 = 15.
Thus the coefficient of x^(-12) is -1 * 15 = -15, as required.
Using the "powers of x" argument again, x^2 must come from x^8 * x^(-6), x^2 * x^0, or x^(-4) * x^6.
x^8 comes from r=2 in the first expansion, so its coefficient is 5C2 * (-1)^2 = 10 * 1 = 10.
x^-6 comes from r=6 in the second expansion, so its coefficient is 6C6 * (-1)^6 = 1 * 1 = 1.
Thus this gives a contribution of 10*1 = 10.
x^2 comes from r=3 in the first expansion, so its coefficient is 5C3 * (-1)^3 = -10 * 1 = -10.
x^0 comes from r=3 in the second expansion, so its coefficient is 6C3 * (-1)^3 = 20 * (-1) = -20.
Thus this gives a contribution of -10 * -20 = 200.
x^(-4) comes from r=4 in the first expansion, so its coefficient is 5C4 * (-1)^4 = 5 * 1 = 5.
x^6 comes from r=0 in the second expansion, so its coefficient is 6C0 * (-1)^0 = 1 * 1 = 1.
Thus this gives a contribution of 5 * 1 = 5.
So the total is 10 + 200 + 5 = 215.
Now you should be able to do the second part, using the hint.
3. (Original post by HapaxOromenon3)
The general term for (x^4 - 1/x^2)^5 is 5Cr * (x^4)^(5-r) * (-1/x^2)^r = 5Cr * x^(20-4r) * (-1)^r * x^(-2r) = 5Cr * (-1)^r * x^(20-6r), for r = 0, 1, 2, 3, 4, 5.
Thus the powers of x will be 20, 14, 8, 2, -4, and -10.
Similarly the general term for (x - 1/x)^6 is 6Cr * x^(6-r) * (-1/x)^r = 6Cr * (-1)^r * x^(6-2r), for r = 0, 1, 2, 3, 4, 5, 6.
Thus the powers of x will be 6, 4, 2, 0 (constant), -2, -4, and -6.
Now notice that x^(-12) must therefore come from x^(-10) * x^(-2). x^(-10) comes from r=5 in the first expansion, so its coefficient is 5C5 * (-1)^5, which is 1 * -1 = -1. x^(-2) comes from r=4 in the second expansion, so its coefficient is 6C4 * (-1)^4 = 15 * 1 = 15.
Thus the coefficient of x^(-12) is -1 * 15 = -15, as required.
Using the "powers of x" argument again, x^2 must come from x^8 * x^(-6), x^2 * x^0, or x^(-4) * x^6.
x^8 comes from r=2 in the first expansion, so its coefficient is 5C2 * (-1)^2 = 10 * 1 = 10.
x^-6 comes from r=6 in the second expansion, so its coefficient is 6C6 * (-1)^6 = 1 * 1 = 1.
Thus this gives a contribution of 10*1 = 10.
x^2 comes from r=3 in the first expansion, so its coefficient is 5C3 * (-1)^3 = -10 * 1 = -10.
x^0 comes from r=3 in the second expansion, so its coefficient is 6C3 * (-1)^3 = 20 * (-1) = -20.
Thus this gives a contribution of -10 * -20 = 200.
x^(-4) comes from r=4 in the first expansion, so its coefficient is 5C4 * (-1)^4 = 5 * 1 = 5.
x^6 comes from r=0 in the second expansion, so its coefficient is 6C0 * (-1)^0 = 1 * 1 = 1.
Thus this gives a contribution of 5 * 1 = 5.
So the total is 10 + 200 + 5 = 215.
Now you should be able to do the second part, using the hint.
Thanks a lot I was completely baffled but now I understand it better.

Updated: November 14, 2016
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