f(x)=sqrt(1+x^3) Use the binomial series to find the Maclaurin series for the function.

A binomial series is an example of infinite series. It is a series that is only convergent when we have |x|lt1 and with a sum of (1+x)^k  where k is any number. To apply binomial series in determining the Maclaurin series of a given function f(x) = (1+x)^k , we may apply the formula:
(1+x)^k = sum_(n=0)^oo (k(k-1)(k-2)...(k-n+1))/(n!) x ^n
 or
(1+x)^k = 1+kx +(k(k-1))/(2!)x^2+(k(k-1)(k-2))/(3!)x^3+(k(k-1)(k-2)(k-3))/(4!)x^4+...
To evaluate the given function f(x) = sqrt(1+x^3) , we may apply the radical property: sqrt(x)= x^(1/2) . The function becomes:
f(x) =(1+x^3)^(1/2)
or
f(x) =(1+x^3)^0.5
To apply the aforementioned formula for binomial series, we may replace "x " with "x^3 " and "k" with "0.5 ". We let:
(1+x^3)^0.5 = sum_(n=0)^oo (0.5(0.5-1)(0.5-2)...(0.5-n+1))/(n!) (x^3) ^n
=sum_(n=0)^oo (0.5(-0.5)(-1.5)...(0.5-n+1))/(n!) x^(3n)
=1+0.5x^(3*1) +(0.5(-0.5))/(2!)x^(3*2)+(0.5(-0.5)(-1.5))/(3!)x^(3*3)+(0.5(-0.5)(-1.5)(-2.5))/(4!)x^(3*4)+...
=1+0.5x^3-0.25/(1*2)x^6+0.375/(1*2*3)x^9-0.9375/(1*2*3*4)x^(12)+...
=1+0.5x^3-0.25/2x^6+0.375/6x^9-0.9375/24x^(12)+...
=1+x^3/2-x^6/8+x^9/16-(5x^(12))/128+...
Then, the Maclaurin series for the f(x)=sqrt(1+x^3) can be expressed as:
sqrt(1+x^3)=1+x^3/2-x^6/8+x^9/16-(5x^(12))/128+...

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