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 +...`