If |alpha|

Question

(C) The given series are infinite geometric series with common ratios $-\alpha$ and $-\beta$ respectively.$s_1 = \frac{1}{1 - (-\alpha)} = \frac{1}{1

Vedclass · Accepted Answer

$\frac{s_1s_2}{1 - s_1 - s_2 + 2s_1s_2}$

Vedclass · Answer

(C) The given series are infinite geometric series with common ratios $-\alpha$ and $-\beta$ respectively.$s_1 = \frac{1}{1 - (-\alpha)} = \frac{1}{1 + \alpha} \implies 1 + \alpha = \frac{1}{s_1} \implies \alpha = \frac{1}{s_1} - 1 = \frac{1 - s_1}{s_1}$.Similarly,$s_2 = \frac{1}{1 + \beta} \implies \beta = \frac{1 - s_2}{s_2}$.Let $S = 1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots \infty$. This is an infinite geometric series with common ratio $-\alpha\beta$.$S = \frac{1}{1 - (-\alpha\beta)} = \frac{1}{1 + \alpha\beta}$.Substituting the values of $\alpha$ and $\beta$:$S = \frac{1}{1 + (\frac{1 - s_1}{s_1})(\frac{1 - s_2}{s_2})} = \frac{1}{1 + \frac{(1 - s_1)(1 - s_2)}{s_1s_2}}$.$S = \frac{s_1s_2}{s_1s_2 + (1 - s_1 - s_2 + s_1s_2)} = \frac{s_1s_2}{2s_1s_2 - s_1 - s_2 + 1}$.

If $|\alpha| < 1$ and $|\beta| < 1$,where $s_1 = 1 - \alpha + \alpha^2 - \alpha^3 + \dots \infty$ and $s_2 = 1 - \beta + \beta^2 - \beta^3 + \dots \infty$,then $1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots \infty$ equals:

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