If f(x) = 1 + x + x^2 + ldots + x^{1000},then | vedclass

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(B) Given the function $f(x) = 1 + x + x^2 + \ldots + x^{1000}$.This is a geometric series with $1001$ terms,where the first term $a = 1$ and the comm

Vedclass · Accepted Answer

$-500$

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(B) Given the function $f(x) = 1 + x + x^2 + \ldots + x^{1000}$.This is a geometric series with $1001$ terms,where the first term $a = 1$ and the common ratio $r = x$.The sum of the series is $f(x) = \frac{x^{1001} - 1}{x - 1}$ for $x 
eq 1$.To find $f^{\prime}(x)$,we use the quotient rule: $\left( \frac{u}{v} ight)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$.Here $u = x^{1001} - 1$ and $v = x - 1$,so $u^{\prime} = 1001x^{1000}$ and $v^{\prime} = 1$.$f^{\prime}(x) = \frac{(1001x^{1000})(x - 1) - (x^{1001} - 1)(1)}{(x - 1)^2}$.Now,substitute $x = -1$:$f^{\prime}(-1) = \frac{(1001(-1)^{1000})(-1 - 1) - ((-1)^{1001} - 1)}{(-1 - 1)^2}$.$f^{\prime}(-1) = \frac{(1001 	imes 1)(-2) - (-1 - 1)}{(-2)^2}$.$f^{\prime}(-1) = \frac{-2002 - (-2)}{4} = \frac{-2002 + 2}{4} = \frac{-2000}{4} = -500$.

If $f(x) = 1 + x + x^2 + \ldots + x^{1000}$,then $f^{\prime}(-1) = $ . . . . . .

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