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

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Answer 1 · 2017-02-20T07:30:23

The answer is #=1+1/3x^2-1/9x^4+5/162x^6+......#

The binomial series is

#(1+b)^(alpha)=sum_(k=0) ^ oo((alpha),(k))b^k#

#=1+alphab+(alpha(alpha-1))/(2!)b^2+(alpha(alpha-1)(alpha-2))/(3!)b^3+.......#

Here,

we have

#b=x^2#

#alpha=1/3#

#(1+x^2)^(1/3)=1+1/3x^2+(1/3*-2/3)/(1*2)x^4+(1/3*-2/3*-5/6)/(1*2*3)x^6+..#

#=1+1/3x^2-1/9x^4+5/162x^6+......#