A) \[^{n}{{C}_{4}}\]
B) \[^{n}{{C}_{4}}{{+}^{n}}{{C}_{2}}\]
C) \[^{n}{{C}_{4}}{{+}^{n}}{{C}_{2}}{{+}^{n}}{{C}_{4}}{{\cdot }^{n}}{{C}_{2}}\]
D) \[^{n}{{C}_{4}}{{+}^{n}}{{C}_{2}}{{+}^{n}}{{C}_{1}}{{\cdot }^{n}}{{C}_{2}}\]
Correct Answer: D
Solution :
\[{{(1+x+{{x}^{2}}+{{x}^{3}})}^{n}}={{(1+x)}^{n}}{{(1+{{x}^{2}})}^{n}}\] \[=(1{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+...{{+}^{n}}{{C}_{n}}{{x}^{n}})\] \[\times (1{{+}^{n}}{{C}_{1}}{{x}^{2}}{{+}^{n}}{{C}_{2}}{{x}^{4}}+...{{+}^{n}}{{C}_{n}}{{x}^{2n}})\] \[\therefore \]The coefficient of\[{{x}^{4}}\] \[{{=}^{n}}{{C}_{2}}{{+}^{n}}{{C}_{2}}{{\cdot }^{n}}{{C}_{1}}{{+}^{n}}{{C}_{4}}\] \[{{=}^{n}}{{C}_{4}}{{+}^{n}}{{C}_{2}}{{+}^{n}}{{C}_{1}}{{\cdot }^{n}}{{C}_{1}}\]
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