A) \[^{n}{{C}_{4}}\]
B) \[^{n}{{C}_{4}}{{+}^{n}}{{C}_{2}}\]
C) \[^{n}{{C}_{4}}+{{\,}^{n}}{{C}_{2}}+\,{{\,}^{n}}{{C}_{4}}{{.}^{n}}{{C}_{2}}\]
D) \[^{n}{{C}_{4}}+{{\,}^{n}}{{C}_{2}}+{{\,}^{n}}{{C}_{1}}.{{\,}^{n}}{{C}_{2}}\]
Correct Answer: D
Solution :
\[{{(1+x+{{x}^{2}}+{{x}^{3}})}^{n}}=\left\{ {{(1+x)}^{n}}{{(1+{{x}^{2}})}^{n}} \right\}\] \[=(1+{{\,}^{n}}{{C}_{1}}x+{{\,}^{n}}{{C}_{2}}{{x}^{2}}+....+{{\,}^{n}}{{C}_{n}}{{x}^{n}})\] \[(1+{{\,}^{n}}{{C}_{1}}{{x}^{2}}+{{\,}^{n}}{{C}_{2}}{{x}^{4}}+....+{{\,}^{n}}{{C}_{n}}{{x}^{2n}})\] Therefore the coefficient of x 4 = \[^{n}{{C}_{2}}+{{\,}^{n}}{{C}_{2}}.{{\,}^{n}}{{C}_{1}}+{{\,}^{n}}{{C}_{4}}\]= \[^{n}{{C}_{4}}+{{\,}^{n}}{{C}_{2}}+{{\,}^{n}}{{C}_{1}}.{{\,}^{n}}{{C}_{2}}\]
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