Q. If $|x| < 1,\$ then the sum of the series $1 + 2x + 3x^{2} + 4x^{3} + ...\infty$ will be
A) $\frac{1}{1 - x}$
B) $\frac{1}{1 + x}$
C) $\frac{1}{(1 + x^{2})}$
D) $\frac{1}{(1 - x)^{2}}$
Explanation
(d)
Let $S = 1 + 2x + 3x^{2} + 4x^{3} + ...\infty$ …(i)
$xS = x + 2x^{2} + 3x^{3} + ...\infty$ …(ii)
Subtracting Eq. (ii) from Eq. (i), we get
$(1 - x)S = 1 + x + x^{2} + x^{3} + ...\infty$
$\Rightarrow \ S = \frac{1}{(1 - x)}\left( \frac{1}{1 - x} \right) = \frac{1}{(1 - x)^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.