Binomial Theorem and Expansion

[0313phd]

This problem asks: What is the middle term in the expansion of (x- 1/x)to the sixth power.

The first thing I thought of was Pascal's triangle which would make the middle term 20, but the correct answer is negative twenty. I am not sure how to expand this difference of squares, using the Binomial Theorem. I know that the first term coefficient is one because 6!/5!1! is 1, which gives a coefficient of one, to x to the sixth power. I am not sure how to do the rest of it. Thanks,
0313

 

[Denis]

6C3 x^3(-1/x)^3 = 20x^3(-1/x^3) = -20

 

[JeffM]

To expand on Denis's explanation a bit, when you expand a difference instead of a sum, the expansion that represents the coefficient of any odd power of the term with a minus sign will retain the minus sign. Do you see why? Pascal's triangle still works if you are careful about plus and minus signs because they are not now all plus.

 

[0313phd]

I am sorry, but I still don't understand the binomial exansion of the difference of the sums
0313

 

[JeffM]

I am sorry, but I still don't understand the binomial exansion of the difference of the sums
0313

Let's do a few

(a - b)[sup:2aro61d9]0[/sup:2aro61d9] = 1.
(a - b)[sup:2aro61d9]1[/sup:2aro61d9] = a - b = (1)a + (-1)b.
(a - b)[sup:2aro61d9]2[/sup:2aro61d9] = a[sup:2aro61d9]2[/sup:2aro61d9] -2ab + b[sup:2aro61d9]2[/sup:2aro61d9] = (1)a[sup:2aro61d9]2[/sup:2aro61d9] + (-2)ab + (1)b[sup:2aro61d9]2[/sup:2aro61d9].
(a - b)[sup:2aro61d9]3[/sup:2aro61d9] = (a[sup:2aro61d9]2[/sup:2aro61d9] - 2ab + b[sup:2aro61d9]2[/sup:2aro61d9])(a - b) = a[sup:2aro61d9]3[/sup:2aro61d9] - ba[sup:2aro61d9]2[/sup:2aro61d9] - 2ba[sup:2aro61d9]2[/sup:2aro61d9] + 2ab[sup:2aro61d9]2[/sup:2aro61d9] + 2ab[sup:2aro61d9]2[/sup:2aro61d9] - b[sup:2aro61d9]3[/sup:2aro61d9] = (1)a[sup:2aro61d9]3[/sup:2aro61d9] + (-3)ba[sup:2aro61d9]2[/sup:2aro61d9] + (3)ab[sup:2aro61d9]2[/sup:2aro61d9] + (-1)b[sup:2aro61d9]3[/sup:2aro61d9].

Note that the absolute values of the coefficients are the same as in Pascal's triangle, but the signs differ. They alternate between plus and minus.
The reason is that when you multiply a plus and a minus you get a minus. So if you are expanding (a - b)[sup:2aro61d9]n[/sup:2aro61d9] every term that has an odd power of b has a negative coefficient.

Make sense now?

 

[0313phd]

Yes, I understand now. Thank you very much. 0313