Consider first the general form of the binomial theorem. If we have (x + y)n, the full product will have terms of the form xny0, xn-1y1, xn-2y2, ..., x1yn-1, x0yn.
The coefficients of these terms will then be C(n,0), C(n,1), C(n,2), ..., C(n,n-1), C(n,n), where C is the "choose" function.
In our case, y is -(x-1⁄2). If we number the terms in the expansion from 0 to n=7, the kth term will be C(7,k) x7-k(-(x-1⁄2))k. This simplifies to (-1)kC(7,k) x(7-k) - k⁄2 = (-1)kC(7,k) x7- 3k⁄2.
We want the term in which the exponent of x is -1&frasl2. So we want 7 - 3k⁄2 = -1&frasl2, or 14 - 3k = -1, or 3k = 15, or k = 5.
The coefficient is then (-1)5C(7,5) = -7!⁄(5!2!) = -7*6/2 = -21.
I just expanded using the 7th line of pascals triangle. :) It probably took about the same amount of time.
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