GO Classes Test Series 2024 | Mock GATE | Test 14 | Question: 56

Question

The coefficient of $x^6$ in the expansion of $A(x)$ is, where
$$
A(x)=\frac{x(1+x)}{(1-x)^3}
$$

Answer 1

We have
$$
A(x)=\frac{x(1+x)}{(1-x)^3}=\left(x+x^2\right)(1-x)^{-3}=\left(x+x^2\right) \sum_{n \geq 0}\left(\begin{array}{c}
-3 \\
n
\end{array}\right)(-x)^n
$$

Hence the coefficient of $x^n$ in $A(x)$ is
$$
n^2, \quad \text { for } n \geq 0 .
$$

Answer 2

$ A(x) = x(1+x)(1-x)^{-3}$
Expanding it we get => $x(1-x)^{-3} + x^{2}(1-x)^{-3}$
We need $x^6$

So Answer will be Coeff($x^5$) in $(1-x)^{-3}$  + Coeff($x^4$) in $(1-x)^{-3}$

 Coeff($x^5$) in $(1-x)^{-3}$ => $\binom{-3}{5} (-1)^5 = (-1)^5 \binom{3+5-1}{5} (-1)^5 =\binom{7}{5} = 21$

 Coeff($x^4$) in $(1-x)^{-3}$ => $\binom{-3}{4} (-1)^4$= $(-1)^4\binom{-3}{4} (-1)^4= \binom{3+4-1}{4}=\binom{6}{4} = 15$

$21+15=36$