A) 476
B) 496
C) 506
D) 528
Correct Answer: A
Solution :
[a] We rewrite the given expression as \[{{[1+{{x}^{2}}(1-x)]}^{8}}\] and expand by using the binomial theorem. We have, \[{{[1+{{x}^{2}}(1-x)]}^{8}}{{=}^{8}}{{C}_{0}}{{+}^{8}}{{C}_{1}}{{x}^{2}}(1-x){{+}^{8}}{{C}_{2}}{{x}^{4}}{{(1-x)}^{2}}\]\[{{+}^{8}}{{C}_{3}}{{x}^{6}}{{(1-x)}^{3}}{{+}^{8}}{{C}_{4}}{{x}^{8}}{{(1-x)}^{4}}\] \[{{+}^{8}}{{C}_{5}}{{x}^{10}}{{(1-x)}^{5}}+...\] The two terms which contain \[{{x}^{10}}\] are \[^{8}{{C}_{4}}\]\[{{x}^{8}}{{(1-x)}^{8}}\]and \[^{8}{{C}_{5}}{{x}^{10}}{{(1-x)}^{5}}\]. Thus, the coefficient of \[{{x}^{10}}\]in the given expression is given by \[^{8}{{C}_{4}}\][Coefficient of \[{{x}^{2}}\] in the expansion of \[{{(1-x)}^{4}}+{{[}^{8}}{{C}_{5}}\] \[{{=}^{8}}{{C}_{4}}(6){{+}^{8}}{{C}_{5}}=\frac{8!}{4!4!}(6)+\frac{8!}{3!5!}\] \[=(70)(6)+56=476\]
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