Arithmetico-geometric Progression

Question

Determine the sum to \infty of the arithmetico-geometric series 1 + 2x + 3x^2 + 4x^3 + ... when the variable x satisfies the convergence condition |x| < 1.

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An arithmetico-geometric series multiplies an arithmetic factor by a geometric factor termwise, and JEE Advanced sums it using the shift-and-subtract technique. Let S = 1 + 2x + 3x^2 + 4x^3 + ..., then xS = x + 2x^2 + 3x^3 + .... Subtracting, S - xS = 1 + x + x^2 + x^3 + ..., whose right side is the geometric series summing to 1/(1 - x) for |x| < 1. Therefore S(1 - x) = 1/(1 - x), giving S = 1/(1 - x)^2. Option 1/(1 - x) ignores the arithmetic coefficients entirely. Option x/(1 - x)^2 is the sum of x + 2x^2 + ..., off by a factor of x. Option 1/(1 + x)^2 wrongly flips the sign of the ratio. Hence S = 1/(1 - x)^2. The result is also the derivative of the geometric series 1/(1 - x) with respect to x, which explains why differentiating a generating function reproduces exactly these linearly increasing coefficients, a viewpoint JEE Advanced rewards for cross-checking. Plausibility check: at x = 0 the series collapses to its first term 1, and 1/(1 - 0)^2 = 1, matching the boundary case exactly.

Arithmetico-geometric Progression

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