The sum of 1 + frac{2}{5} + frac{3}{5^2} + fr | vedclass

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Question

(A) Let the sum be $S_n = 1 + \frac{2}{5} + \frac{3}{5^2} + \frac{4}{5^3} + \dots + \frac{n}{5^{n-1}}$.Multiply by $\frac{1}{5}$:$\frac{1}{5}S_n = \fr

Vedclass · Accepted Answer

$\frac{25}{16} - \frac{4n + 5}{16 	imes 5^{n-1}}$

Vedclass · Answer

(A) Let the sum be $S_n = 1 + \frac{2}{5} + \frac{3}{5^2} + \frac{4}{5^3} + \dots + \frac{n}{5^{n-1}}$.Multiply by $\frac{1}{5}$:$\frac{1}{5}S_n = \frac{1}{5} + \frac{2}{5^2} + \frac{3}{5^3} + \dots + \frac{n}{5^n}$.Subtracting the two equations:$(1 - \frac{1}{5})S_n = 1 + (\frac{2}{5} - \frac{1}{5}) + (\frac{3}{5^2} - \frac{2}{5^2}) + \dots + (\frac{n}{5^{n-1}} - \frac{n-1}{5^{n-1}}) - \frac{n}{5^n}$.$\frac{4}{5}S_n = 1 + \frac{1}{5} + \frac{1}{5^2} + \dots + \frac{1}{5^{n-1}} - \frac{n}{5^n}$.The sum of the geometric progression is $\frac{1(1 - (1/5)^n)}{1 - 1/5} = \frac{1 - 1/5^n}{4/5} = \frac{5}{4}(1 - \frac{1}{5^n})$.$\frac{4}{5}S_n = \frac{5}{4}(1 - \frac{1}{5^n}) - \frac{n}{5^n}$.$S_n = \frac{25}{16}(1 - \frac{1}{5^n}) - \frac{5n}{4 	imes 5^n} = \frac{25}{16} - \frac{25}{16 	imes 5^n} - \frac{20n}{16 	imes 5^n}$.$S_n = \frac{25}{16} - \frac{25 + 20n}{16 	imes 5^n} = \frac{25}{16} - \frac{5(5 + 4n)}{16 	imes 5 	imes 5^{n-1}} = \frac{25}{16} - \frac{4n + 5}{16 	imes 5^{n-1}}$.

The sum of $1 + \frac{2}{5} + \frac{3}{5^2} + \frac{4}{5^3} + \dots$ up to $n$ terms is

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