Question

Find the coefficient of $\frac{1}{x}$ in $(1+3x^3+x^4){(1+\frac{1}{x})}^8$.

Answer 1

$(1+3x^3+x^4)(1+\frac{1}{x}{)}^8$

$(a+b{)}^n={}^n{C}_0{a}^n{b}^0{+}^n{C}_1{a}^{n-1}{b}^1+---{+}^n{C}_n{a}^o{b}^n$

${(1+\frac{1}{x})}^8={}^8{C}_0(1{)}^8{\left(\frac{1}{x}\right)}^0+{}^8{C}_1(1{)}^7{\left(\frac{1}{x}\right)}^1+{}^8{C}_2(1{)}^6{\left(\frac{1}{x}\right)}^2+{}^8{C}_3(1{)}^5{\left(\frac{1}{x}\right)}^3+{}^8{C}_4(1{)}^4{\left(\frac{1}{x}\right)}^4+{}^{}{}^8{C}_5(1{)}^3{\left(\frac{1}{x}\right)}^5$

${+}^8{C}_6{\left(\frac{1}{x}\right)}^6+{}^8{C}_7{\left(\frac{1}{x}\right)}^7+{}^8{C}_8{\left(\frac{1}{x}\right)}^8$

$=1+\frac{8}{x}+\frac{28}{x^2}+\frac{56}{x^3}+\frac{70}{x^4}+\frac{56}{x^5}+\frac{28}{x^6}+\frac{8}{x^7}+\frac{1}{x^8}$

$\therefore (1+3x^3+x^4)(1+\frac{8}{x}+\frac{28}{x^2}+\frac{56}{x^3}+\frac{70}{x^4}+\frac{56}{x^5}+\frac{28}{x^6}+\frac{8}{x^7}+\frac{1}{x^8})$

coefficient of $\left(\frac{1}{x}\right)$

$=1\times 8+3\times 70+56\times 1$

$=274$