A) \[-1\]
B) \[-1/2\]
C) \[0\]
D) \[1/2\]
Correct Answer: B
Solution :
\[{{(1-x-2{{x}^{2}})}^{6}}={{(1+x)}^{6}}{{(1-2x)}^{6}}\]\[=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....+{{a}_{12}}{{x}^{12}}\] Putting \[x=1/2\], we have \[0=1+{{a}_{1}}/2+{{a}_{2}}/{{2}^{2}}+{{a}_{3}}/{{2}^{3}}+\]\[{{a}_{4}}/{{2}^{4}}+...+{{a}_{12}}/{{2}^{2}}\] ? (1) Putting \[x=-1/2\], we have \[1=1-{{a}_{1}}/2+{{a}_{2}}/{{2}^{2}}-{{a}_{3}}/{{2}^{3}}+{{a}_{4}}/{{2}^{4}}-...{{a}_{12}}/{{2}^{12}}\] ? (2) Adding (1) and (2), we have \[1=2(1+{{a}_{2}}/{{2}^{2}}+{{a}_{4}}/{{2}^{4}}+...+{{a}_{12}}/{{2}^{12}})\] \[\Rightarrow \]\[{{a}_{2}}/{{2}^{2}}+{{a}_{4}}/{{2}^{4}}+{{a}_{6}}/{{2}^{6}}+...+{{a}_{12}}/{{2}^{12}}=-1/2\]You need to login to perform this action.
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