Find coefficient of x^n in expansion of (1-x)^0.5. Show it is equal to (2n)!/(2^2n)(n!)^2Thanks in advance =)

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beckden | High School Teacher | (Level 1) Educator

Posted on

The binomial theorem

`(a+b)^n` = `sum_(k=0)^n((n),(k))a^(n-k) b^k`

when n is not an integer




So `a = 1` , `b = -x` and `n=-1/2`





noting that `(1*3*5*...*(2k-1))=((2k)!)/((2^k)k!)` so the general form is




So our answer is