6 votes 6 votes The coefficient of $x^6$ in the expansion of $A(x)$ is, where$$A(x)=\frac{x(1+x)}{(1-x)^3}$$ Combinatory goclasses2024-mockgate-14 numerical-answers combinatory recurrence-relation 2-marks + – GO Classes asked Feb 5 • recategorized Feb 5 by Lakshman Bhaiya GO Classes 586 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes 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 $$ n^2, \quad \text { for } n \geq 0 . $$ GO Classes answered Feb 5 GO Classes comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes $ 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 $(1-x)^{-3}$ => $\binom{-3}{4} (-1)^4$= $(-1)^4\binom{-3}{4} (-1)^4= \binom{3+4-1}{4}=\binom{6}{4} = 15$ $21+15=36$ squirrel69 answered Feb 6 squirrel69 comment Share Follow See all 0 reply Please log in or register to add a comment.