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

$b=x^2$

$alpha=1/3$

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

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

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