Expand the following expression in ascending powers of x as far as ${x^3}$ for$\dfrac{{1 + 2x}}{{1 - x - {x^2}}}$.

Hint: In this question remember to compare given expression with a general representation of polynomial expression which is given as ${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$, use this information to approach the solution of the question.

Complete step-by-step answer:
We need to expand $\dfrac{{1 + 2x}}{{1 - x - {x^2}}}$ in ascending powers of x as far as ${x^3}$ only.
Now let $\dfrac{{1 + 2x}}{{1 - x - {x^2}}} = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$
So, taking denominator to the LHS part we can say that
$1 + 2x = (1 - x - {x^2})({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...)$
So, we get
$1 + 2x = ({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...) - (x{a_0} + {x^2}{a_1} + {x^3}{a_2} + ...) - ({x^2}{a_0} + {x^3}{a_1} + {x^4}{a_2} + ...$
Now we will be comparing the coefficients of x with x and ${x^2}$ with ${x^2}$ and so on, for both the LHS and the RHS part but only up to${x^3}$.
On comparing the coefficients, we get
${a_0} = 1,{a_1} - {a_0} = 2{\text{ }}$
So, putting values we get
${{\text{a}}_1} = 3$
Now the coefficients of higher power of x are found in succession form and following the relation ${a_n} - {a_{n - 1}} - {a_{n - 2}} = 0$
Hence put n = 2
${a_2} - {a_{2 - 1}} - {a_{2 - 2}} = 0$
$ \Rightarrow $${a_2} - {a_1} - {a_0} = 0$
Putting the values, we get ${a_2} - 4 = 0 \Rightarrow {a_2} = 4$
Now putting n = 3
${a_3} - {a_{3 - 1}} - {a_{3 - 2}} = 0$
$ \Rightarrow $${a_3} - {a_2} - {a_1} = 0$
Putting the values, we get ${a_3} - 7 = 0 \Rightarrow {a_3} = 7$
Thus, let’s substitute the value back to obtain the desired expansion we get
$\dfrac{{1 + 2x}}{{1 - x - {x^2}}} = 1 + 3x + 4{x^2} + 7{x^3} + ...$

Note: The key approach to solve such type of problem statements is always to equate the given expression with a general polynomial equation. Then using the information provided in the question simply evaluate to obtain the desired coefficients, putting these values back will eventually give the expanded expression.