The sum of the series 1 + 2x + 3x^2 + 4x^3 + | vedclass

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(A) Let $S_n$ be the sum of the given series to $n$ terms:$S_n = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n - 1}$ $(i)$Multiply by $x$:$xS_n = x + 2x^2 + 3x

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$\frac{1 - (n + 1)x^n + nx^{n + 1}}{(1 - x)^2}$

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(A) Let $S_n$ be the sum of the given series to $n$ terms:$S_n = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n - 1}$ $(i)$Multiply by $x$:$xS_n = x + 2x^2 + 3x^3 + \dots + (n - 1)x^{n - 1} + nx^n$ $(ii)$Subtracting $(ii)$ from $(i)$:$(1 - x)S_n = 1 + x + x^2 + x^3 + \dots + x^{n - 1} - nx^n$Using the sum formula for a geometric progression with $n$ terms:$(1 - x)S_n = \frac{1 - x^n}{1 - x} - nx^n$$(1 - x)S_n = \frac{1 - x^n - nx^n(1 - x)}{1 - x}$$(1 - x)S_n = \frac{1 - x^n - nx^n + nx^{n + 1}}{1 - x}$$(1 - x)S_n = \frac{1 - (n + 1)x^n + nx^{n + 1}}{1 - x}$Therefore,$S_n = \frac{1 - (n + 1)x^n + nx^{n + 1}}{(1 - x)^2}$.

The sum of the series $1 + 2x + 3x^2 + 4x^3 + \dots$ up to $n$ terms is:

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