A) \[32\]
B) \[63\]
C) \[64\]
D) \[31\]
Correct Answer: D
Solution :
\[{{(1+x-2{{x}^{2}})}^{6}}=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{12}}{{x}^{12}}\] On putting \[x=1\] and \[x=-1\] and adding the results, \[64=2(1+{{a}_{2}}+{{a}_{4}}+...)\] \[\Rightarrow \] \[{{a}_{2}}+{{a}_{4}}+{{a}_{6}}+...+{{a}_{12}}=31\]You need to login to perform this action.
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