If |alpha|

Question

(C) Given the infinite geometric series $s_1 = 1 - \alpha + \alpha^2 - \alpha^3 + \dots = \frac{1}{1 + \alpha}$ and $s_2 = 1 - \beta + \beta^2 - \beta

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the infinite geometric series $s_1 = 1 - \alpha + \alpha^2 - \alpha^3 + \dots = \frac{1}{1 + \alpha}$ and $s_2 = 1 - \beta + \beta^2 - \beta^3 + \dots = \frac{1}{1 + \beta}$.From these,we have $1 + \alpha = \frac{1}{s_1} \implies \alpha = \frac{1}{s_1} - 1$ and $1 + \beta = \frac{1}{s_2} \implies \beta = \frac{1}{s_2} - 1$.Let $s = 1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots = \frac{1}{1 + \alpha\beta}$.Substituting the values of $\alpha$ and $\beta$:$s = \frac{1}{1 + (\frac{1}{s_1} - 1)(\frac{1}{s_2} - 1)} = \frac{1}{1 + (\frac{1 - s_1}{s_1})(\frac{1 - s_2}{s_2})} = \frac{s_1s_2}{s_1s_2 + (1 - s_1)(1 - s_2)}$.$s = \frac{s_1s_2}{s_1s_2 + 1 - s_1 - s_2 + s_1s_2} = \frac{s_1s_2}{1 - s_1 - s_2 + 2s_1s_2}$.

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

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