Question

The coefficient of $x^4$ in the expansion of ${(1-x+2x^2)}^{12}$ is:

• A. ${}^{12}{C}_3$
• B. ${}^{13}{C}_3$
• C. ${}^{14}{C}_4$
• D. ${}^{12}{C}_3+3{}^{13}{C}_3+{}^{14}{C}_4$

Answer 1

The correct option is D ${}^{12}{C}_3+3{}^{13}{C}_3+{}^{14}{C}_4$

${(1-x+2x^2)}^{12}$

$={(1+x(2x-1\left)\right)}^{12}$

$={{}^{12}C}_0{1}^{12}+{{}^{12}C}_1{1}^{11}{\left(x\right(2x-1\left)\right)}^1+{{}^{12}C}_2{1}^{10}{\left(x\right(2x-1\left)\right)}^2+{{}^{12}C}_3{1}^9{\left(x\right(2x-1\left)\right)}^3+...+{{}^{12}C}_{12}{1}^{12}{\left(x\right(2x-1\left)\right)}^{12}$

Coefficient of $x^4=0$ in second and third term

Third term $={{}^{12}C}_2{1}^{10}{\left(x\right(2x-1\left)\right)}^2={{}^{12}C}_2{x}^2(4x^2-4x+1)={{}^{12}C}_2(4x^4-4x^3+x^2)$

Thus,coefficient of $x^4=4{{}^{12}C}_2$

Fourth term $={{}^{12}C}_3(8x^6-x^3-12x^5+6x^4)$ (on expansion)

$\therefore$ Coefficient of $x^4$ is $6{{}^{12}C}_3$

Fifth term $={{}^{12}C}_4{x}^4[{{}^{4}C}_0{\left(2x\right)}^4+{{}^{4}C}_1{\left(2x\right)}^3(-1)+{{}^{4}C}_2{\left(2x\right)}^2{(-1)}^2+{{}^{4}C}_3(2x){(-1)}^3+{{}^{4}C}_4{(-1)}^4]$ (on expansion)

$={{}^{12}C}_4[{{}^{4}C}_{0}16x^8-{{}^{4}C}_{1}8x^7+{{}^{4}C}_{2}4x^6+{{}^{4}C}_{3}2x^5+x^4]$

$\therefore$ Coefficient of $x^4$ is ${{}^{12}C}_4$

Sixth term $={{}^{12}C}_5\left({\left(x\right(2x-1\left)\right)}^5\right)$ has no $x^4$ term when multiplied by$x^5$

Using the formula ${{}^{n+1}C}_r={{}^{n}C}_r+{{}^{n}C}_{r-1}$

Adding all the coefficients of $x^4=4{{}^{12}C}_2+6{{}^{12}C}_3+{{}^{12}C}_4$

$=4({{}^{12}C}_2+{{}^{12}C}_3)+2{{}^{12}C}_3+{{}^{12}C}_4$ (by rearranging)

$=4\left({{}^{13}C}_3\right)+{{}^{12}C}_3+({{}^{12}C}_4+{{}^{12}C}_3)$

$=4{{}^{13}C}_3+{{}^{12}C}_3+{{}^{13}C}_4$

$=3{{}^{13}C}_3+({{}^{13}C}_3+{{}^{13}C}_4)+{{}^{12}C}_3$

$=3{{}^{13}C}_3+{{}^{14}C}_4+{{}^{12}C}_3$