If we expand the expression $\frac{1+2x}{1-x-x^2}$ in ascending powers of $x$ as far as $x^3$, we get:

Given:
The expression $\frac{1+2x}{1-x-x^2}$

To expand the given expression in ascending powers of $x$ as far as $x^3$ assume that

$\begin{array}{rcl}\frac{1+2x}{1-x-x^2}& =& a_0+a_{1}x+a_2{x}^2+a_3{x}^3+........\\ 1+2x& =& \left(1-x-x^2\right)\left(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+........\right)\\ & =& \left(a_0+a_{1}x+a_2{x}^2+a_3{x}^2+....\right)-\left(a_{0}x+a_1{x}^2+a_2{x}^3+....\right)-\left(a_0{x}^2+a_1{x}^3+...\right)\end{array}$

Now, compare the coefficients of $x$ with $x$ ,  $x^2$ with $x^2$ and $x^3$ with $x^3$ for both sides of the equation.

So,

$a_0=1$ and

$\begin{array}{rcl}{a}_1-a_0& =& 2\\ a_1-1& =& 2\\ a_1& =& 3\end{array}$

The coefficients of higher power of $x$ are in succession form and thus,

$a_n-a_{n-1}-a_{n-2}=0$

Let $n=2$, so

$\begin{array}{rcl}{a}_2-a_{2-1}-a_{2-2}& =& 0\\ a_2-a_1-a_0& =& 0\\ a_2-3-1& =& 0\\ a_2& =& 4\end{array}$
 

Now, let $n=3$, so

$\begin{array}{rcl}{a}_3-a_{3-1}-a_{3-2}& =& 0\\ a_3-a_2-a_1& =& 0\\ a_3-4-3& =& 0\\ a_2& =& 7\end{array}$

Therefore, the expansion of the expression $\frac{1+2x}{1-x-x^2}$ in ascending powers of $x$ as far as $x^3$ will be $\begin{array}{rcl} & & 1+3x+4x^2+7x^3+........\end{array}$.

Hence, the correct answer is option A.