A) \[31\]
B) \[32\]
C) \[64\]
D) \[1024\]
Correct Answer: A
Solution :
We have\[{{(1+x-2{{x}^{2}})}^{6}}=1+{{a}_{1}}x\] \[+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+....+{{a}_{12}}{{x}^{12}}\] ... (i) On putting \[x=1\] and \[x=-1\] respectively in Eq. (i), we get \[{{(1+1-2)}^{6}}=1+{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{12}}\] \[\Rightarrow \] \[0=1+{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{12}}\] ? (ii) and\[{{(1-1-2)}^{6}}=1-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}+....+{{a}_{12}}\] \[\Rightarrow \] \[64=1-{{a}_{1}}+{{a}_{2}}-{{a}_{3}}+....+{{a}_{12}}\] ? (iii) On adding Eqs. (ii) and (iii), we get \[64=2(1+{{a}_{2}}+{{a}_{4}}+....+{{a}_{12}})\] \[\Rightarrow \] \[{{a}_{2}}+{{a}_{4}}+{{a}_{6}}+...+{{a}_{12}}=31\]
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