Find the sum S_n = 1 + 2x + 3x^2 + 4x^3 + dot | vedclass

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(A) Given the series $S_n = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n-1} \dots (1)$Multiply by $x$: $xS_n = x + 2x^2 + 3x^3 + \dots + (n-1)x^{n-1} + nx^n \

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

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(A) Given the series $S_n = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n-1} \dots (1)$Multiply by $x$: $xS_n = x + 2x^2 + 3x^3 + \dots + (n-1)x^{n-1} + nx^n \dots (2)$Subtracting $(2)$ from $(1)$:$(1 - x)S_n = 1 + (2x - x) + (3x^2 - 2x^2) + \dots + (nx^{n-1} - (n-1)x^{n-1}) - nx^n$$(1 - x)S_n = 1 + x + x^2 + \dots + x^{n-1} - nx^n$The sum of the geometric series $1 + x + x^2 + \dots + x^{n-1}$ is $\frac{1 - x^n}{1 - x}$.So,$(1 - x)S_n = \frac{1 - x^n}{1 - x} - nx^n$$(1 - x)S_n = \frac{1 - x^n - nx^n(1 - x)}{1 - x}$$S_n = \frac{1 - x^n - nx^n + nx^{n+1}}{(1 - x)^2}$$S_n = \frac{1 - (n + 1)x^n + nx^{n+1}}{(1 - x)^2}$

Find the sum $S_n = 1 + 2x + 3x^2 + 4x^3 + \dots$ up to $n$ terms.

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