A) 32
B) 63
C) 64
D) None of these
Correct Answer: D
Solution :
\[{{(1+x-2{{x}^{2}})}^{6}}=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+....+{{a}_{12}}{{x}^{12}}\]. Putting \[x=1\] and x = ? 1 and adding the results 64 = 2(1+a2+a4+...) \[\therefore \,\,\,{{a}_{2}}+{{a}_{4}}+{{a}_{6}}+....+{{a}_{12}}=31\].
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