question_answer

The sum of the co-efficient of all even degree terms in x in the expansion of \[{{\left( x+\sqrt{{{x}^{3}}-1} \right)}^{6}}+\]\[{{\left( x-\sqrt{{{x}^{3}}-1} \right)}^{6}},(x>1)\]is equal to : [JEE Main 8-4-2019 Morning]

A) 32                               

B) 26

C) 29                               

D) 24

Correct Answer: D

Solution :

\[{{\left( x-\sqrt{{{x}^{3}}-1} \right)}^{6}}+{{\left( x-\sqrt{{{x}^{3}}-1} \right)}^{6}}\]           \[{{+}^{6}}{{C}_{6}}{{({{x}^{3}}-1)}^{3}}]\]           \[=2{{[}^{6}}{{C}_{0}}{{x}^{6}}+{}^{6}{{C}_{2}}{{x}^{7}}-{}^{6}{{C}_{2}}{{x}^{4}}+{}^{6}{{C}_{4}}{{x}^{8}}+{}^{6}{{C}_{4}}{{x}^{2}}\] \[-2\,\,{}^{6}{{C}_{4}}{{x}^{5}}+({{x}^{9}}-1-3{{x}^{6}}+3{{x}^{3}})]\] \[\Rightarrow \]Sum of coefficient of even powers of x \[=2[115+15+1513]=24\]