Expand me generally

Using a binomial expansion for $\sqrt { 1+2x } or\quad { (1+2x) }^{ \frac { 1 }{ 2 } }$

The general binomial ${ (1+a) }^{ n }=1\ +\frac { n }{ 1! } a+\frac { n(n-1) }{ 2! } { a }^{ 2 }+\frac { n(n-1)(n-2) }{ 3! } { a }^{ 3 }+...$

by letting a = 2x and n = 1/2

we obtain $1+\frac { 1 }{ 2 } \left( 2x \right) +\frac { \frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } \right) }{ 2! } { (2x) }^{ 2 }+\frac { \frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } \right) \left( -\frac { 3 }{ 2 } \right) }{ 3! } { \left( 2x \right) }^{ 3 }+...$

Simplifying to $1+x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 2 } +...$

let $x=\frac { 1 }{ 4 }$ ... well within the convergence interval

We obtain x = 1.2267 rounding to 1.227

I was not looking for solvers to find the value by direct input into the calculator using the square root function.