1+2x+3x2+4x3+........∞≥4 , then

S=1+2x+3x2+4x3+........∞ ⇒ xS=x+2x2+3x3+4x4+.......∞ ⇒ (1−x)S=1+x+x2+.......∞=11−x ⇒ S=1(1−x)2 Now, S≥4 ⇒ 1(1−x)2>4 ⇒ (x−1)2≤14 ⇒ −12≤x−1≤12 ⇒ 12≤x≤32 Also 0<|x|<1 ⇒ 12≤x<1

1+2x+3x2+4x3+........∞≥4 , then

S=1+2x+3x2+4x3+........∞ ⇒ xS=x+2x2+3x3+4x4+.......∞ ⇒ (1−x)S=1+x+x2+.......∞=11−x ⇒ S=1(1−x)2 Now, S≥4 ⇒ 1(1−x)2>4 ⇒ (x−1)2≤14 ⇒ −12≤x−1≤12 ⇒ 12≤x≤32 Also 0<|x|<1 ⇒ 12≤x<1

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$1 + 2 x + 3 x^{2} + 4 x^{3} + ........ \infty \geq 4$ , then

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least value of x is $\frac{1}{2}$

greatest value of x is $\frac{4}{3}$

least value of x is $\frac{2}{3}$

greatest value of x is 2

answer is A.

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Detailed Solution

$S = 1 + 2 x + 3 x^{2} + 4 x^{3} + ........ \infty \Rightarrow x S = x + 2 x^{2} + 3 x^{3} + 4 x^{4} + ....... \infty \Rightarrow (1 - x) S = 1 + x + x^{2} + ....... \infty = \frac{1}{1 - x} \Rightarrow S = \frac{1}{{(1 - x)}^{2}} N o w, S \geq 4 \Rightarrow \frac{1}{{(1 - x)}^{2}} > 4 \Rightarrow {(x - 1)}^{2} \leq \frac{1}{4} \Rightarrow - \frac{1}{2} \leq x - 1 \leq \frac{1}{2} \Rightarrow \frac{1}{2} \leq x \leq \frac{3}{2} A l s o 0 < | x | < 1 \Rightarrow \frac{1}{2} \leq x < 1$