To show that the equation

$\frac{1}{(1-x{)}^{2}}=1+2x+3{x}^{2}+\dots +(n+1){x}^{n}+\frac{n+2-(n+1)x}{(1-x{)}^{2}}{x}^{(n+1)}$

is true, we'll begin by expanding the right side of the equation and then simplifying it step by step.

Let's start with the expansion:

$1+2x+3{x}^{2}+\dots +(n+1){x}^{n}+\frac{n+2-(n+1)x}{(1-x{)}^{2}}{x}^{(n+1)}$

To find a common denominator for the terms involving $x$, we can multiply each term by $(1-x{)}^{2}$:

$(1-x{)}^{2}+2x(1-x{)}^{2}+3{x}^{2}(1-x{)}^{2}+\dots +(n+1){x}^{n}(1-x{)}^{2}+(n+2-(n+1)x){x}^{(n+1)}$

Now, let's simplify each term:

$(1-x{)}^{2}=(1-2x+{x}^{2})$

$2x(1-x{)}^{2}=2x(1-2x+{x}^{2})=2x-4{x}^{2}+2{x}^{3}$

$3{x}^{2}(1-x{)}^{2}=3{x}^{2}(1-2x+{x}^{2})=3{x}^{2}-6{x}^{3}+3{x}^{4}$

Continuing this pattern, we can simplify the remaining terms:

$(n+1){x}^{n}(1-x{)}^{2}=(n+1){x}^{n}-2(n+1){x}^{n+1}+(n+1){x}^{n+2}$

$(n+2-(n+1)x){x}^{(n+1)}=(n+2){x}^{(n+1)}-(n+1){x}^{(n+2)}$

Now, let's combine all the simplified terms:

$(1-2x+{x}^{2})+(2x-4{x}^{2}+2{x}^{3})+(3{x}^{2}-6{x}^{3}+3{x}^{4})+\dots +((n+1){x}^{n}-2(n+1){x}^{n+1}+(n+1){x}^{n+2})+((n+2){x}^{(n+1)}-(n+1){x}^{(n+2)})$

We can see that the common terms cancel each other out, leaving only the outermost terms:

$1+(n+2){x}^{(n+1)}-(n+1){x}^{(n+2)}$

Now, let's simplify the terms further:

$1+(n+2){x}^{(n+1)}-(n+1){x}^{(n+2)}=\frac{1}{(1-x{)}^{2}}+\frac{(n+2){x}^{(n+1)}}{(1-x{)}^{2}}-\frac{(n+1){x}^{(n+2)}}{(1-x{)}^{2}}$

Since all terms have a common denominator, we can combine them into a single fraction:

$\frac{1}{(1-x{)}^{2}}+\frac{(n+2){x}^{(n+1)}}{(1-x{)}^{2}}-\frac{(n+1){x}^{(n+2)}}{(1-x{)}^{2}}=\frac{1+(n+2){x}^{(n+1)}-(n+1){x}^{(n+2)}}{(1-x{)}^{2}}$

Finally, we can see that the numerator of the fraction is the same as the right side of the original equation. Therefore, we have:

$\frac{1}{(1-x{)}^{2}}=1+2x+3{x}^{2}+\dots +(n+1){x}^{n}+\frac{n+2-(n+1)x}{(1-x{)}^{2}}{x}^{(n+1)}$

And that concludes the proof.