A) \[\frac{{{a}^{2}}}{{{(1-a)}^{2}}}-\frac{{{b}^{2}}}{{{(1-b)}^{2}}}\]
B) \[\frac{{{a}^{2}}-{{b}^{2}}}{(1-{{a}^{2}})(1-{{b}^{2}})}\]
C) \[\frac{{{a}^{2}}}{{{(1-a)}^{2}}}+\frac{{{b}^{2}}}{{{(1-b)}^{2}}}\]
D) \[\frac{{{a}^{2}}}{{{(1+a)}^{2}}}-\frac{{{b}^{2}}}{{{(1+b)}^{2}}}\]
E) \[\frac{a}{1+a}-\frac{b}{1+b}\]
Correct Answer: B
Solution :
\[\because \]\[\det \,({{A}_{1}})=\left| \begin{matrix} a & b \\ b & a \\ \end{matrix} \right|={{a}^{2}}-{{b}^{2}}\] \[\det \,({{A}_{2}})=\left| \begin{matrix} {{a}^{2}} & {{b}^{2}} \\ {{b}^{2}} & {{a}^{2}} \\ \end{matrix} \right|={{a}^{4}}-{{b}^{4}}\] \[\therefore \]\[\sum\limits_{i=1}^{\infty }{\det }({{A}_{i}})=\det ({{A}_{1}})+\det ({{A}_{2}})+....\] \[={{a}^{2}}-{{b}^{2}}+{{a}^{4}}-{{b}^{4}}+.....\] \[=\frac{{{a}^{2}}}{1-{{a}^{2}}}-\frac{{{b}^{2}}}{1-{{b}^{2}}}\] \[=\frac{{{a}^{2}}-{{a}^{2}}{{b}^{2}}-{{b}^{2}}+{{a}^{2}}{{b}^{2}}}{(1-{{a}^{2}})(1-{{b}^{2}})}\]
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