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

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Question

$5+55+555+\ldots$

Let $S_{n}=5+55+555+\ldots .$ to $n$ terms

$=\frac{5}{9}[9+99+999+\ldots \ldots 	ext { to } n 	ext { terms }]$

$=\frac{5}

Vedclass · Accepted Answer

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

Vedclass · Answer

$5+55+555+\ldots$

Let $S_{n}=5+55+555+\ldots .$ to $n$ terms

$=\frac{5}{9}[9+99+999+\ldots \ldots 	ext { to } n 	ext { terms }]$

$=\frac{5}{9}\left[(10-1)+\left(10^{2}-1ight)+\left(10^{3}-1ight)+\ldots 	ext { to } n 	ext { terms }ight]$

$=\frac{5}{9}[\left(10+10^{2}+10^{3}+	ext { to } n 	ext { terms }ight)$

$-(1+1+\ldots 	ext { to } n 	ext { terms })]$

$=\frac{5}{9}\left[\frac{10\left(10^{n}-1ight)}{10-1}-night]$

$=\frac{5}{9}\left[\frac{10\left(10^{n}-1ight)}{9}-night]$

$=\frac{50}{81}\left(10^{n}-1ight)-\frac{5 n}{9}$

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: $5+55+555+\ldots$