A) \[{{3}^{2010}}\]
B) 1
C) \[{{2}^{2010}}\]
D) none of these
Correct Answer: C
Solution :
[c] put \[x=\omega ,{{\omega }^{2}}\] \[{{(3+\omega +{{\omega }^{2}})}^{2010}}={{a}_{0}}+{{a}_{1}}\omega +{{a}_{2}}{{\omega }^{2}}+...\] \[\Rightarrow {{2}^{2010}}={{a}_{0}}+{{a}_{1}}{{\omega }^{2}}+{{a}_{2}}\omega +{{a}_{3}}+{{a}_{2}}\omega +...\] And \[{{2}^{2010}}={{a}_{0}}+{{a}_{1}}{{\omega }^{2}}+{{a}_{2}}\omega +{{a}_{3}}+{{a}_{2}}\omega +...\] Adding (1) and (2), we have \[2\times {{2}^{2010}}=2{{a}_{0}}-{{a}_{1}}-{{a}_{2}}+2{{a}_{3}}-{{a}_{4}}-{{a}_{5}}+2{{a}_{6}}-...\] Or \[{{2}^{2010}}={{a}_{0}}-\frac{1}{2}{{a}_{1}}-\frac{1}{2}{{a}_{2}}+{{a}_{3}}-\frac{1}{2}{{a}_{4}}-\frac{1}{2}{{a}_{5}}+{{a}_{6...}}\]
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