If (20)^{19} + 2(21)(20)^{18} + 3(21)^2(20)^{ | vedclass

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(C) Let $S = (20)^{19} + 2(21)(20)^{18} + 3(21)^2(20)^{17} + \ldots + 20(21)^{19}$.Dividing by $(20)^{19}$,we get $k = 1 + 2(\frac{21}{20}) + 3(\frac{

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$400$

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(C) Let $S = (20)^{19} + 2(21)(20)^{18} + 3(21)^2(20)^{17} + \ldots + 20(21)^{19}$.Dividing by $(20)^{19}$,we get $k = 1 + 2(\frac{21}{20}) + 3(\frac{21}{20})^2 + \ldots + 20(\frac{21}{20})^{19}$.Let $x = \frac{21}{20}$. Then $k = 1 + 2x + 3x^2 + \ldots + 20x^{19}$.Multiplying by $x$,we get $kx = x + 2x^2 + 3x^3 + \ldots + 20x^{20}$.Subtracting the two equations: $k(1 - x) = 1 + x + x^2 + \ldots + x^{19} - 20x^{20}$.Using the sum of a geometric progression: $k(1 - x) = \frac{1 - x^{20}}{1 - x} - 20x^{20}$.Since $1 - x = 1 - \frac{21}{20} = -\frac{1}{20}$,we have $k(-\frac{1}{20}) = \frac{1 - x^{20}}{-1/20} - 20x^{20} = -20(1 - x^{20}) - 20x^{20}$.$k(-\frac{1}{20}) = -20 + 20x^{20} - 20x^{20} = -20$.Therefore,$k = (-20) 	imes (-20) = 400$.

If $(20)^{19} + 2(21)(20)^{18} + 3(21)^2(20)^{17} + \ldots + 20(21)^{19} = k (20)^{19}$,then $k$ is equal to

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