If y = 1 + x + x^2 + x^3 + . . . + infty and | vedclass

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(A) Given the infinite geometric series $y = 1 + x + x^2 + x^3 + . . . + \infty$ for $|x| < 1$. Using the sum formula for an infinite geometric series

Vedclass · Accepted Answer

$2 y y^{\prime}$

Vedclass · Answer

(A) Given the infinite geometric series $y = 1 + x + x^2 + x^3 + . . . + \infty$ for $|x| Using the sum formula for an infinite geometric series,$y = \frac{1}{1 - x}$. Differentiating with respect to $x$,we get $y^{\prime} = \frac{d}{dx}(1 - x)^{-1} = -1(1 - x)^{-2}(-1) = \frac{1}{(1 - x)^2}$. Now,differentiating $y^{\prime}$ with respect to $x$ to find $y^{\prime \prime}$,we get $y^{\prime \prime} = \frac{d}{dx}(1 - x)^{-2} = -2(1 - x)^{-3}(-1) = \frac{2}{(1 - x)^3}$. We can rewrite this as $y^{\prime \prime} = 2 	imes \frac{1}{(1 - x)} 	imes \frac{1}{(1 - x)^2}$. Since $y = \frac{1}{1 - x}$ and $y^{\prime} = \frac{1}{(1 - x)^2}$,we have $y^{\prime \prime} = 2 y y^{\prime}$.

If $y = 1 + x + x^2 + x^3 + . . . + \infty$ and $|x| < 1$,then $y^{\prime \prime} =$

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