Find the generating function and the general term of the following series: $2 + 3x + 5{x^2} + 9{x^3} + ....$

Hint: Multiply the given series by $x$ and then from the original series which is given in the question. Observe the pattern after subtracting the series and use the binomial expansion of ${(1 - x)^n}$ to evaluate the generating function.

Complete step-by-step answer:
We are given a series $2 + 3x + 5{x^2} + 9{x^3} + ....$
We have to find the generating function and the general term of the given series.
Generating functions are used to represent the terms of an infinite series to a simple form.
Let the given series is $P$.
Therefore, according to the question $P = 2 + 3x + 5{x^2} + 9{x^3} + ....$
Multiply both sides by $x$ in the given series.
$Px = 2x + 3{x^2} + 5{x^3} + 9{x^4} + .....$
Subtract the obtained series after multiplying it by $x$ from the given series.
That is, $P - Px = 2 + 3x + 5{x^2} + 9{x^3}... - (2x + 3{x^2} + 5{x^3} + 9{x^4})$
Solve the series.
$
  P(1 - x) = 2 + 3x - 2x + 5{x^2} - 3{x^2} + 9{x^3} - 5{x^3}.... \\
   \Rightarrow P(1 - x) = 2 + x + 2{x^2} + 4{x^3} + .... \\
 $
Factor out the common factor $x$ from the given series to obtain some pattern.
$P(1 - x) = 2 + x(1 + 2x + 4{x^2} + ....).....(1)$
Now we try to write the series in the bracket in the binomial form so that we can get our generating function.
For this, we use the binomial expansion formula.
We know the formula for ${(1 - x)^n}$.
Use the formula.
\[{(1 - x)^n} = 1 - nx + \dfrac{{n(n - 1)}}{{2!}}{x^2} - \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + .....\]
Substitute $x = 2x$ and $n = - 1$ in the above formula.
$
  {(1 - 2x)^{ - 1}} = 1 - ( - 1)2x + \dfrac{{( - 1)( - 1 - 1)}}{{2!}}{(2x)^2} + ... \\
   \Rightarrow {(1 - 2x)^{ - 1}} = 1 + 2x + \dfrac{2}{{2 \times 1}} + 4{x^2} + ... \\
   \Rightarrow {(1 - 2x)^{ - 1}} = 1 + 2x + 4{x^2} + ... \\
 $
Hence, the equation $(1)$ can be written as:
$P(1 - x) = 2 + x{(1 - 2x)^{ - 1}}$
Use the property \[{a^{ - x}} = \dfrac{1}{{{a^x}}}\]to simplify the expression.
 $
  P(1 - x) = 2 + \dfrac{x}{{1 - 2x}} \\
   \Rightarrow P(1 - x) = \dfrac{{2 - 4x + x}}{{1 - 2x}} \\
   \Rightarrow P = \dfrac{{2 - 3x}}{{(1 - 2x)(1 - x)}} \\
 $
Therefore, the generating function of the given series is $\dfrac{{2 - 3x}}{{(1 - 2x)(1 - x)}}$.
Rewrite the given series to evaluate the general term.
$P = ({x^0} + {(2x)^0}) + ({x^1} + {(2x)^1}) + ({x^2} + {(2x)^2}) + ({x^3} + {(2x)^3}) + ....$
Observing the above pattern, the general term for the series can be written as:
${x^n} + {(2x)^n} = {x^n}(1 + {2^n})$

$\therefore$ the general term for the series ${x^n}(1 + {2^n})$, here $n = 0,1,2,3.....$

Note:
There is another method by which we can solve the series $1 + 2x + 4{x^2} + ....$which is show below:
The given series $1 + 2x + 4{x^2} + ....$ is a G.P. series with common ratio $2x$.
We know the sum of the infinite G.P. series which is $S = \dfrac{a}{{1 - r}}$, here $a$ is the first term and $r$ is the common ratio.
In our series, $a = 1$ and $r = 2x$
Substitute the values in the formula and evaluate the sum.
$S = \dfrac{1}{{1 - 2x}} = {(1 - 2x)^{ - 1}}$
Therefore, $P(1 - x) = 2 + x{(1 - 2x)^{ - 1}}$
Use the property \[{a^{ - x}} = \dfrac{1}{{{a^x}}}\]to simplify the expression.
 $
  P(1 - x) = 2 + \dfrac{x}{{1 - 2x}} \\
   \Rightarrow P(1 - x) = \dfrac{{2 - 4x + x}}{{1 - 2x}} \\
   \Rightarrow P = \dfrac{{2 - 3x}}{{(1 - 2x)(1 - x)}} \\
 $
Therefore, the generating function of the given series is $\dfrac{{2 - 3x}}{{(1 - 2x)(1 - x)}}$.