A) \[\frac{{{3}^{10}}+1}{2}\]
B) \[\frac{{{3}^{9}}+1}{2}\]
C) \[\frac{{{3}^{10}}-1}{2}\]
D) \[\frac{{{3}^{9}}-1}{2}\]
E) \[{{2}^{19}}-1\]
Correct Answer: C
Solution :
\[\because \]\[{{(1+x-3{{x}^{2}})}^{10}}=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}\] \[+....+{{a}_{20}}{{x}^{20}}\] On putting\[x=-1,\]we get \[{{(1-1-3)}^{10}}=1-{{a}_{1}}+{{a}_{2}}-.....+{{a}_{20}}\] \[\Rightarrow \] \[{{3}^{10}}=1-{{a}_{1}}+{{a}_{2}}-....+{{a}_{20}}\] ?..(i) Now, on putting\[x=1,\]we get \[{{(1+1-3)}^{10}}=1+{{a}_{1}}+{{a}_{2}}+....+{{a}_{20}}\] \[\Rightarrow \] \[{{3}^{10}}=1-{{a}_{1}}+{{a}_{2}}+....+{{a}_{20}}\] ?.(ii) From Eqs. (i) and (ii), we get \[{{3}^{10}}+1=2(1+{{a}_{2}}+{{a}_{4}}+...+{{a}_{20}})\] \[\Rightarrow \] \[\frac{{{3}^{10}}-1}{2}={{a}_{2}}+{{a}_{4}}+.....+{{a}_{20}}\]
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