Discussion
The sum of the series \(1 + 2x + 3x^2 + 4x^3 + ...\) for \(|x|<1\) is:
Answer: C
This is an Arithmetico-Geometric Progression (A.G.P.) with the A.P. part being 1, 2, 3, ... and G.P. part being \(1, x, x^2, ...\). The sum of an infinite A.G.P. is given by \(S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}\). Here, the first term of the A.P. is \(a_{AP}=1\), \(d=1\), and \(r=x\). The first term of the A.G.P. is \(a=1\). The sum is \(S = \frac{1}{1-x} + \frac{1 \cdot x}{(1-x)^2} = \frac{1-x+x}{(1-x)^2} = \frac{1}{(1-x)^2}\). This is a standard result for the expansion of \((1-x)^{-2}\).
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