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# How do you use the binomial series to expand the function ##f(x)=(1-x)^(2/3)## ?

This is a natural extension of raising a binomial to a whole number power: ##(a+b)^n=sum_(k=0)^n (n/k)a^(n-k)b^k## (sorry about the inner bar) Where ##(n/k)=(n!)/(k!(n-k)!)=(n(n-1)…(n-k+1))/(k!)##

So we can apply this to any exponent r even if r is an arbitrary real number.

##(a+b)^r=sum_(k=0)^oo (r(r-1)…(r-k+1))/(k!) a^(r-k)b^k##

Now put your info in, with ##r=2/3, a=1, b=-x:##

##(1-x)^(2/3)=sum_(k=0)^oo (2/3(2/3-1)…(2/3-k+1))/(k!) 1^(2/3-k)(-x)^k##

##=(1)^(2/3)+(2/3)/1(1)^(-1/3)(-x)^1+((2/3)(2/3-1))/2(1)^(-4/3)(-x)^2+…##

##=1-2/3x-1/9x^2+…##

There's the start of the series; I dare you to compute the next two terms. Take the \dansmath challenge/!