The binomial theorem

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

when n is not an integer

`((n),(k))=((n)(n-1)(n-2)...(n-k+1))/(k!)`

and

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

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

`(1-x)^(-1/2)=sum_(k=0)^oo((-1/2),(k))1^(n-k)(-x)^k=sum_(k=0)^oo((-1/2),(k))(-1)^kx^k`

`((-1/2),(0))=(1)/(0!)=1`

`((-1/2),(1))=((-1/2))/(1!)=-1/2`

`((-1/2),(2))=((-1/2)(-3/2))/(2!)=-(1*3)/(2^2(2!))`

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

`((-1/2),(k))=(-1)^k((2k)!)/(2^kk!2^kk!)`

so:

`(1-x)^(-1/2)=sum_(k=0)^oo(-1)^k((2k)!)/(2^(2k)(k!)^2)(-1)^kx^k`

So our answer is

`(1-x)^(-1/2)=sum_(k=0)^oo((2k)!)/(2^(2k)(k!)^2)x^k`

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The binomial theorem

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

when n is not an integer

`((n),(k))=((n)(n-1)(n-2)...(n-k+1))/(k!)`

and

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

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

`(1-x)^(-1/2)=sum_(k=0)^oo((-1/2),(k))1^(n-k)(-x)^k=sum_(k=0)^oo((-1/2),(k))(-1)^kx^k`

`((-1/2),(0))=(1)/(0!)=1`

`((-1/2),(1))=((-1/2))/(1!)=-1/2`

`((-1/2),(2))=((-1/2)(-3/2))/(2!)=-(1*3)/(2^2(2!))`

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

`((-1/2),(k))=(-1)^k((2k)!)/(2^kk!2^kk!)`

so:

`(1-x)^(-1/2)=sum_(k=0)^oo(-1)^k((2k)!)/(2^(2k)(k!)^2)(-1)^kx^k`

So our answer is

`(1-x)^(-1/2)=sum_(k=0)^oo((2k)!)/(2^(2k)(k!)^2)x^k`