This answer is for those who don’t know how to solve Arithmetico–Geometric series.
$\sum_{n=0}^{\infty} x^n = 1+x + x^2 + x^3 +…… = \frac{1}{1-x}$ $(1)$
Differentiating with respect to $x$
$n \sum_{n=0}^{\infty}x^{n-1} = \frac{1}{(1-x)^2}$
$n \sum_{n=1}^{\infty}x^{n-1} = \frac{1}{(1-x)^2}$
So, $<1,2,3,…..>\;\; \leftrightarrow \;\;\frac{1}{(1-x)^2}$ $(2)$
Replace $x$ with $x^2$ in $(1)$
$\sum_{n=0}^{\infty} x^{2n} = \frac{1}{1-x^2}$
$\sum_{n=0}^{\infty} <1,1,…..> x^{2n} = \frac{1}{1-x^2}$ $(3)$
Suppose, sequence $\{g_n\}$ has a generating function is $G(z).$ i.e. $\sum_{n}^{}g_n z^n = G(z),$ So,
$G(z) + G(-z) = \sum_{n}^{}g_n (1+(-1)^n) z^n$
It means, $\frac{G(z) + G(-z)}{2} = \sum_{n}^{}g_{2n} z^{2n}$ (Formula to get even-numbered sequence i.e. $g_0,g_2,g_4,..$)
Similarly, $\frac{G(z) - G(-z)}{2} = \sum_{n}^{}g_{2n+1} z^{2n+1}$ (Formula to get odd-numbered sequence i.e. $g_1,g_3,g_5,..$)
So, from $(2),$ we get the odd-numbered sequence,
$ \sum_{n=0}^{\infty}<2,4,6,….> x^{2n+1}= \frac{\frac{1}{(1-x)^2} – \frac{1}{(1+x)^2}}{2}$ $(4)$
Now, comes to given sequence in the question,
$<1,2,1,4,1,6,….>\;\; \leftrightarrow \;\; G(x)$
Applying the formula to get even and odd numbered sequence and from $(4)$ and $(3)$
$\frac{G(x) + G(-x)}{2} = \frac{1}{1-x^2}$
$\frac{G(x) – G(-x)}{2} = \frac{\frac{1}{(1-x)^2} – \frac{1}{(1+x)^2}}{2}$
Adding these $2$ equations to eliminate $G(-x)$,
$$G(x) = \frac{1}{1-x^2} + \frac{\frac{1}{(1-x)^2} – \frac{1}{(1+x)^2}}{2}$$
To get the correct option, we need to simplify it which I will not do.
Since, $G(x)$ is simply a formula, so, I can put $x=0,1,2,..$ (For $x \neq 1$, $G(x)$ is not defined ) and check options which does match with our $G(x)$ since all options have $\frac{1}{1-x}$ and $(1 – x^2)^2$ in denominator of first part, we would have to compute it once.
For, $x=0$ all options give $G(0) = 1$ and our $G(0)=1$ also.
Now, for $x=2,$ our not good-looking $G(x)$ gives, $G(2) = \frac{1}{9}$
In options, $G(2)$ in $ A) \frac{10}{9} – 1 = \frac{1}{9}$, $(B) \frac{-2}{9} – 1 = -ve,$ $(C) \frac{4}{9} – 1 = -ve,$ $(D) \frac{2}{9} – 1 = -ve$
Hence, Answer is $(A)$
