If |x|

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(C) We know that $(1 - x)^{-1} = 1 + x + x^2 + x^3 + . . . \infty$. Differentiating with respect to $x$,we get $(1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 +

Vedclass · Accepted Answer

$76$

Vedclass · Answer

(C) We know that $(1 - x)^{-1} = 1 + x + x^2 + x^3 + . . . \infty$. Differentiating with respect to $x$,we get $(1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + . . . \infty$. Differentiating again with respect to $x$,we get $2(1 - x)^{-3} = 1 \cdot 2 + 2 \cdot 3x + 3 \cdot 4x^2 + . . . \infty$. Thus,$(1 - x)^{-3} = \frac{1}{2}(1 \cdot 2 + 2 \cdot 3x + 3 \cdot 4x^2 + . . . \infty)$. Substituting this into the given expression: $[\frac{1}{2}(1 \cdot 2 + 2 \cdot 3x + 3 \cdot 4x^2 + . . . \infty)]^{-25} = [(1 - x)^{-3}]^{-25} = (1 - x)^{75}$. The expansion of $(1 - x)^{75}$ is a finite polynomial with $75 + 1 = 76$ terms.

If $|x| < 1$,then the number of terms in the expansion of $[\frac{1}{2}(1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^2 + . . . . . . \infty)]^{-25}$ is

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