MCQ
${\rm{If}}\,|\alpha |,|\beta |\, < 1,\,\,1 - \alpha + {\alpha ^2} - {\alpha ^3} + ...\infty = {s_1},$ $1 - \beta  + {\beta ^2} - {\beta ^3} + ....\infty  = {s_2},$ then $1 - \alpha \beta + {a^2}{\beta ^2} - {a^3}{\beta ^3} + ....\infty \,\,{\rm{equals}}$
  • A
    ${s_1}{s_2}$
  • B
    $\frac{{{s_1}{s_2}}}{{1 + {s_1}{s_2}}}$
  • $\frac{{{s_1}{s_2}}}{{1 - {s_1} - {s_2} + 2{s_1}{s_2}}}$
  • D
    $\frac{1}{{1 + {s_1}{s_2}}}$

Answer

Correct option: C.
$\frac{{{s_1}{s_2}}}{{1 - {s_1} - {s_2} + 2{s_1}{s_2}}}$
c
(c) ${s_1} = \frac{1}{{1 + \alpha }},{s_2} = \frac{1}{{1 + \beta }}$

Let $s = 1 - \alpha \beta + {\alpha ^2}{\beta ^2}.......$

==> $s = \frac{1}{{1 + \alpha \beta }}$
$\alpha = \frac{1}{{{s_1}}} - 1,\,\,\beta = \frac{1}{{{s_2}}} - 1$;

$\therefore s = \frac{1}{{1 + \left( {\frac{1}{{{s_1}}} - 1} \right)\,\left( {\frac{1}{{{s_2}}} - 1} \right)}}$.

$s = \frac{{{s_1}{s_2}}}{{2{s_1}{s_2} + 1 - {s_1} - {s_2}}}$.

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