Question

In the expansion of${(1+x+x^2+x^3)}^6$, the coefficient of $x^{14}$ is

• A. $130$
• B. $120$
• C. $128$
• D. $125$
• E. $115$

Answer 1

The correct option is B

$120$

Explanation for the correct option:

Finding the coefficient of $x^{14}$:

$$\begin{gathered} {\left[(1+x)+x^2(1+x)\right]}^6 \\ \text{Rewriting the equation, we get} \\ \Rightarrow {\left[(1+x)(1+x^2)\right]}^6 \\ \Rightarrow {(1+x)}^6{(1+x^2)}^6 \end{gathered}$$

Expanding it with the help of binomial expansion

We know that ${\left(x+y\right)}^n={}^{n}C_0{x}^n{y}^0+{}^{n}C_1{x}^{n-1}{y}^1+{}^{n}C_2{x}^{n-2}{y}^2+\dots \dots \dots \dots \dots +{}^{n}C_n{x}^0{y}^n$

$$\begin{gathered} =\left({}^6{C}_0+{}^6{C}_{1}x+{}^6{C}_2{x}^2+{}^6{C}_3{x}^3+{}^6{C}_4{x}^4+{}^6{C}_5{x}^5+{}^6{C}_6{x}^6\right)\left({}^6{C}_0+{}^6{C}_1{x}^2+{}^6{C}_2{x}^4+{}^6{C}_3{x}^6+{}^6{C}_4{x}^8+{}^6{C}_5{x}^{10}+{}^6{C}_6{x}^{12}\right) \\ =\left({}^6{C}_2{}^6{C}_6\right)x^{2+12}+\left({}^6{C}_4{}^6{C}_5\right)x^{4+10}+\left({}^6{C}_6{}^6{C}_4\right)x^{6+8}+.......\left[\text{considering only the terms with}{x}^{14}\right] \\ =\left({}^6{C}_2{}^6{C}_6{+}^6{C}_4{}^6{C}_5{+}^6{C}_6{}^6{C}_4\right)x^{14}+....... \\ =\left(\frac{6!}{2!4!}\frac{6!}{6!0!}+\frac{6!}{4!2!}\frac{6!}{5!1!}+\frac{6!}{6!0!}\frac{6!}{4!2!}\right)x^{14}+....... \\ =\left(\frac{6\times 5\times 4\times 3\times 2\times 1}{1\times 2\times 4\times 3\times 2\times 1}1+\frac{6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 2\times 1}\frac{6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1\times 1}+1\frac{6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 2\times 1}\right) \\ =\left(15+15\left(6\right)+15\right)x^{14} \\ =120x^{14} \end{gathered}$$

Hence, option B is correct.