# Find the sum of the following series up to n terms:

Question:

Find the sum of the following series up to n terms:

(i) 5 + 55 + 555 + …

(ii) .6 +.66 +. 666 +…

Solution:

(i) 5 + 55 + 555 + …

Let $S_{n}=5+55+555+\ldots . .$ to $n$ terms

$=\frac{5}{9}[9+99+999+\ldots$ to $\mathrm{n}$ terms $]$

$=\frac{5}{9}\left[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots\right.$ to $\mathrm{n}$ terms $]$

$=\frac{5}{9}\left[\left(10+10^{2}+10^{3}+\ldots \mathrm{n}\right.\right.$ terms $)-(1+1+\ldots \mathrm{n}$ terms $\left.)\right]$

$=\frac{5}{9}\left[\frac{10\left(10^{\mathrm{n}}-1\right)}{10-1}-\mathrm{n}\right]$

$=\frac{5}{9}\left[\frac{10\left(10^{\mathrm{n}}-1\right)}{9}-\mathrm{n}\right]$

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

(ii) $.6+.66+.666+\ldots$

Let $S_{n}=06 .+0.66+0.666+\ldots$ to $n$ terms

$=6[0.1+0.11+0.111+\ldots$ to $n$ terms $]$

$=\frac{6}{9}[0.9+0.99+0.999+\ldots$ to $\mathrm{n}$ terms $]$

$=\frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^{2}}\right)+\left(1-\frac{1}{10^{3}}\right)+\ldots\right.$ to $n$ terms $]$

$=\frac{2}{3}\left[(1+1+\ldots \mathrm{n}\right.$ terms $)-\frac{1}{10}\left(1+\frac{1}{10}+\frac{1}{10^{2}}+\ldots \mathrm{n}\right.$ terms $\left.)\right]$

$=\frac{2}{3}\left[\mathrm{n}-\frac{1}{10}\left(\frac{1-\left(\frac{1}{10}\right)^{\mathrm{n}}}{1-\frac{1}{10}}\right)\right]$

$=\frac{2}{3} n-\frac{2}{30} \times \frac{10}{9}\left(1-10^{-n}\right)$

$=\frac{2}{3} n-\frac{2}{27}\left(1-10^{-n}\right)$