Find the sum of the following series up to n | vedclass

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(B) Let $S_n = 5 + 55 + 555 + \ldots$ to $n$ terms.$S_n = 5(1 + 11 + 111 + \ldots 	ext{ to } n 	ext{ terms})$$S_n = \frac{5}{9}(9 + 99 + 999 + \ldot

Vedclass · Accepted Answer

$\frac{50}{81}(10^n-1)-\frac{5n}{9}$

Vedclass · Answer

(B) Let $S_n = 5 + 55 + 555 + \ldots$ to $n$ terms.$S_n = 5(1 + 11 + 111 + \ldots 	ext{ to } n 	ext{ terms})$$S_n = \frac{5}{9}(9 + 99 + 999 + \ldots 	ext{ to } n 	ext{ terms})$$S_n = \frac{5}{9}((10-1) + (10^2-1) + (10^3-1) + \ldots + (10^n-1))$$S_n = \frac{5}{9}((10 + 10^2 + 10^3 + \ldots + 10^n) - (1 + 1 + 1 + \ldots + 1))$Using the sum formula for a geometric progression $a(r^n-1)/(r-1)$ where $a=10$ and $r=10$:$S_n = \frac{5}{9}\left(\frac{10(10^n-1)}{10-1} - night)$$S_n = \frac{5}{9}\left(\frac{10(10^n-1)}{9} - night)$$S_n = \frac{50}{81}(10^n-1) - \frac{5n}{9}$

Find the sum of the following series up to $n$ terms:
$5+55+555+\ldots$

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Find the sum of the following series up to $n$ terms:$5+55+555+\ldots$