A) 900
B) 909
C) 990
D) 999
Correct Answer: C
Solution :
We have \[1+x+{{x}^{2}}+{{x}^{3}}=(1+x)(1+{{x}^{2}})\] \[\therefore \]\[{{(1+x+{{x}^{2}}+{{x}^{3}})}^{11}}={{(1+x)}^{11}}{{(1+{{x}^{2}})}^{11}}\] \[=({{\,}^{11}}{{C}_{0}}{{+}^{11}}{{C}_{1}}x{{+}^{11}}{{C}_{2}}{{x}^{2}}{{+}^{11}}{{C}_{3}}{{x}^{2}}+\] \[{{\,}^{11}}{{C}_{4}}{{x}^{4}}+...)\] \[({{\,}^{11}}{{C}_{0}}{{+}^{11}}{{C}_{1}}{{x}^{2}}{{+}^{11}}{{C}_{2}}{{x}^{4}}+...)\] \[\Rightarrow \] Coefficient of \[{{x}^{4}}\]in\[{{(1+x+{{x}^{2}}+{{x}^{3}})}^{11}}=\] Coefficient of \[{{x}^{4}}\]in \[={{\,}^{11}}{{C}_{0}}{{.}^{11}}{{C}_{2}}{{+}^{11}}{{C}_{2}}{{.}^{11}}{{C}_{1}}{{+}^{11}}{{C}_{4}}{{.}^{11}}{{C}_{0}}\] \[=990\]
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