• # question_answer Let${{(1-x-2{{x}^{2}})}^{6}}=1+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{12}}{{x}^{12}}$. Then$\frac{{{a}_{2}}}{{{2}^{2}}}+\frac{{{a}_{4}}}{{{2}^{4}}}+\frac{{{a}_{6}}}{{{2}^{6}}}+...+\frac{{{a}_{12}}}{{{2}^{12}}}$is equal A) $-1$                           B) $-1/2$C) $0$                             D) $1/2$

${{(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$