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

Answer 1

Answer: (B)

120

Answer 2

${{(1+x+{{x}^{2}}+{{x}^{3}})}^{6}}$
$={{(1+x)}^{6}}{{(1+{{x}^{2}})}^{6}}$
$={{(}^{6}}{{C}_{0}}{{+}^{6}}{{C}_{1}}x{{+}^{6}}{{C}_{2}}{{x}^{2}}{{+}^{6}}{{C}_{3}}{{x}^{3}}{{+}^{6}}{{C}_{4}}{{x}^{4}}$ ${{+}^{6}}{{C}_{5}}{{x}^{5}}{{+}^{6}}{{C}_{6}}{{x}^{6}})$ $\times {{(}^{6}}{{C}_{0}}{{+}^{6}}{{C}_{1}}{{x}^{2}}{{+}^{6}}{{C}_{2}}{{x}^{4}}{{+}^{6}}{{C}_{3}}{{x}^{6}}{{+}^{6}}{{C}_{4}}{{x}^{8}}$ ${{+}^{6}}{{C}_{5}}{{x}^{10}}{{+}^{6}}{{C}_{6}}{{x}^{12}})$
$\therefore$ Coefficient of ${{x}^{14}}$ in ${{(1+x+{{x}^{2}}+{{x}^{3}})}^{6}}$ ${{=}^{6}}{{C}_{2}}{{.}^{6}}{{C}_{6}}{{+}^{6}}{{C}_{4}}{{.}^{6}}{{C}_{5}}{{+}^{6}}{{C}_{6}}{{.}^{6}}{{C}_{4}}$
$=\frac{6!}{2!4!}.\frac{6!}{0!6!}+\frac{6!}{4!2!}.\frac{6!}{5!1!}+\frac{6!}{0!6!}.\frac{6!}{2!4!}$
$=15+90+15=120$