# Binomial expansion questionWatch

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Thread starter 2 years ago
#1
Can anyone answer this and explain how you got there as I'm struggling to even start answering this question. 0
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2 years ago
#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.
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Thread starter 2 years ago
#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.
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