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

 Recall binomial series follows: 
(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 given function f(x) =sqrt(1+x) , we may apply radical property: sqrt(x)= x^(1/2) . The function becomes:
f(x) =sqrt(1+x)
        = (1+x)^(1/2)
        or (1+x)^(0.5)
 By comparing "(1+x)^k " with "(1+x)^(0.5) ”, we have the corresponding values:
 x=x and k =0.5
Plug-in the values on the formula for binomial series, we get:
(1+x)^(0.5) =sum_(n=0)^oo (0.5(0.5-1)(0.5-2)...(0.5-n+1))/(n!)x^n
        =1 + 0.5x + (0.5(0.5-1))/(2!) x^2 + (0.5(0.5-1)(0.5-2))/(3!)x^3 +(0.5(0.5-1)(0.5-2)(0.5-3))/(4!)x^4+...
=1 + 1/2x -0.25/(2!) x^2 + 0.375/(3!)x^3 -0.9375/(4!)x^4+...
=1 + x/2 -x^2/8 +x^3/16 -(5x^4)/128 +...
Therefore, the Maclaurin series  for  the function f(x) =sqrt(1+x) can be expressed as:
sqrt(1+x)=1 + x/2 -x^2/8 +x^3/16 -(5x^4)/128 +...