(a) To use the binomial series, we need to change the expansion to the form of (1 + u)n.

So we perform the following steps to get it in the required form.

`sqrt(4+x^2)`

`=(4+x^2)^(1/2)`

`=[4(1+x^2/4)]^(1/2)`

`=4^(1/2)(1+x^2/4)^(1/2)`

`=2(1+x^2/4)^(1/2)`

Hence, if we let the "u" term be `x^2/4` and `n=1/2`, and then substituting in the binomial series,

(1 + u)n = 1 + nu + `(n(n-1))/(2!)u^2` `+(n(n-1)(n-2))/(3!)u^3``+...`

we get:

`2(1+(x^2)/4)^(1/2)`

`~~2[1+(1/2)((x^2)/4)` `+((1/2)(1/2-1))/(2!)((x^2)/4)^2` `+((1/2)(1/2-1)(1/2-2))/(3!)((x^2)/4)^3` `{:+...]`

`=2+x^2/4-x^4/64+x^6/512+...`

(b) We now use the above expression to approximate `sqrt(4.25)`.

If we use `x=0.5`, then `x^2=0.25` and we'll have `4+x^2 = 4.25`.

Substituting `x=0.5` into our expansion for `sqrt(4+x^2)`, we have:

`sqrt(4.25)` `~~2+(0.5)^2/4-(0.5)^4/64+(0.5)^6/512` `~~2.06155395508`

Is it correct?

The calculator value for `sqrt(4.25)` is `2.06155281281`, so we've achieved 5 decimal place accuracy using our binomial series. We'd get a better answer if we were to take more than 4 terms.