Question

Find the coefficient of x${}^{13}$ in the expansion of $(1-x{)}^{5}x(1+x+x^2+x^3{)}^4$

Answer 1

$(1-x{)}^{5}x(1+x+x^2+x^3{)}^4$ sum of $4$ terms of G.P$=1+x+x^2+x^3=1\frac{(1-x^4)}{1-x}$

$=(1-x{)}^{5}x{\left(\frac{1-x^4}{1-x}\right)}^4$

$=(1-x{)}^{5}x\frac{(1-x^4{)}^4}{(1-x{)}^4}$

$=(1-x)x(1-x^4{)}^4$

Coefficient of $x^{13}$ in $(1-x)x(1-x^4{)}^4$

$=$Coefficient of $x^{12}$ in $(1-x)(1-x^4{)}^4$

$=$Coefficient of $x^{12}$ in $(1-x^4{)}^4-x(1-x^4{)}^4$

$=$Coefficient of $x^{12}$ in $(1-x^4{)}^4-$coefficient of $x^{12}$ in $x(1-x^4{)}^4$

$=$Coefficient of $x^{12}$ in $(1-x^4{)}^4-0$ ($x^{12}$ term is not possible in $x(1-x^4{)}^4$)

$={}^4{C}_3(-1{)}^3$ $(\because (1-x^4{)}^4={}^4{C}_0-{}^4{C}_1{x}^4+{}^4{C}_2{x}^8-{}^4{C}_3{x}^{12}+{}^4{C}_4{x}^{16})$

$=-4$

$\therefore$ Coefficient of $(1-x{)}^{5}x(1+x+x^2+x^3{)}^4=-4$.