MadeEasy Test Series 2019: Combinatory- Generating Functions

Question

Let $M(x) = \frac{x^{2018}}{(1-x)^{2019}}$

we define $M(x) = \sum_{r=0}^{\infty}a_{r}x^{r}$ ,then $a_{r}$ is equal to-

$A)\binom{r}{2019}$

$B)\binom{r}{r+2018}$

$C)\binom{r}{2019-r}$

$D)\binom{r}{r-2018}$

Comments

what is this shifting property ?

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https://gateoverflow.in/279053/madeeasy-test-series-full-syllbus

Answer 1

We know that

$\large
\begin{align*}
(1-x)^{-n} &= 1 - (-n)x+ \dfrac{(-n)(-n-1)}{2}x^2-\frac{(-n)(-n-1)(-n-2)}{3*2}x^3+...\\
&= 1+\binom{n}{1}x + \binom{n+1}{2}x^2+\binom{n+2}{3}x^3+...\\
&= \sum_{r \ge 0} \binom{n+r-1}{r}x^r
\end{align*}
\large$

Now we have

$\large
\begin{align*}
\frac{x^{2018}}{(1-x)^{2019}} &= x^{2018}(1-x)^{-2019} \\
&= x^{2018}\sum_{r \ge 0}{\binom{2019+r-1}{r}x^r} \\
&= \sum_{r \ge 0}{\binom{r+2018}{r}x^{2018+r}}\\
&= \sum_{r \ge 2018}{\binom{r}{r-2018}x^r}\\
&= \sum_{r \ge 0}{\binom{r}{r-2018}x^r} \because \binom{n}{r} = 0, r < 0
\end{align*}
\large$

Answer 2

Its Generating Function :