Question

Determine the coefficient of x^3 in the expansion of (1 – x)^5(1 + 1/x)^5

Answer 1

Notice that

(1 - x)⁵ (1 + 1/x)⁵ = ((1 - x) (1 + 1/x))⁵ = (1 - x + 1/x - 1)⁵ = (1/x - x)⁵

Recall the binomial theorem:

$\displaystyle(a+b)^n = \sum_{k=0}^n\binom nk a^{n-k}b^k$

Let a = 1/x, b = -x, and n = 5. Then

$\displaystyle\left(\frac1x-x\right)^5 = \sum_{k=0}^5\binom5k\left(\frac1x\right)^{5-k}(-x)^k = \sum_{k=0}^5 (-1)^k\binom5k x^{2k-5}$

We get an x ³ term for

2k - 5 = 3   ==>   2k = 8   ==>   k = 4

so that the coefficient would be

$(-1)^4\dbinom54 = \boxed{5}$