# Find the sum of the following series:

Question:

Find the sum of the following series:

(i) 5 + 55 + 555 + ... to n terms;

(ii) 7 + 77 + 777 + ... to n terms;

(iii) 9 + 99 + 999 + ... to n terms;

(iv) 0.5 + 0.55 + 0.555 + ... to n terms.

(v) 0.6 + 0.66 + 0.666 + .... to n terms

Solution:

(i) We have,

5 + 55 + 555+ ... n terms

Taking 5 as common:

$S_{n}=5[1+11+111+\ldots n$ terms $]$

$=\frac{5}{9}(9+99+999+\ldots n$ terms $)$

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

$=\frac{5}{9}\left\{\left(10+10^{2}+10^{3}+\ldots+10^{n}\right)\right\}-(1+1+1+1+\ldots n$ times $)$

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

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

$=\frac{5}{81}\left\{10^{n+1}-9 n-10\right\}$

(ii) We have,

7 + 77 + 777 + ... n terms

$\mathrm{S}_{n}=7[1+11+111+\ldots \mathrm{n}$ terms $]$

$=\frac{7}{9}(9+99+999+\ldots \mathrm{n}$ terms $)$

$=\frac{7}{9}\left\{(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots+\left(10^{n}-1\right)\right\}$

$=\frac{7}{9}\left\{\left(10+10^{2}+10^{3}+\ldots+10^{n}\right)\right\}-(1+1+1+1 \ldots n$ times $)$

$=\frac{7}{9}\left\{10 \times \frac{\left(10^{n}-1\right)}{10-1}-n\right\}=\frac{7}{9}\left\{\frac{10}{9}\left(10^{n}-1\right)-n\right\}$

$=\frac{7}{81}\left\{10^{n+1}-9 n-10\right\}$

(iii) We have,

9 + 99 + 999 + ... n terms

$=(9+99+999+\ldots+$ to $n$ terms $)$

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

$=\left\{\left(10+10^{2}+10^{3}+\ldots+10^{n}\right)\right\}-(1+1+1+1 \ldots n$ times $)$

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

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

$=\frac{1}{9}\left\{10^{n+1}-9 n-10\right\}$

(iv) We have,

0.5 + 0.55 + 0.555 + ... n terms

$S_{n}=5[0.1+0.11+0.111+\ldots n$ terms $]$

$=\frac{5}{9}(0.9+0.99+0.999+\ldots+$ to $n$ terms $)$

$=\frac{5}{9}\left\{\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\ldots n\right.$ terms $\}$

$=\frac{5}{9}\left\{\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)+\ldots n\right.$ terms $\}$

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

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

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

(v) We have,

0.6 + 0.66 +.666 + ... to terms

$S_{n}=6[0.1+0.11+0.111+\ldots n$ terms $]$

$=\frac{6}{9}(0.9+0.99+0.999+\ldots n$ terms $)$

$=\frac{6}{9}\left\{\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\ldots n\right.$ terms $\}$

$=\frac{6}{9}\left\{\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)+\ldots n\right.$ terms $\}$

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

$=\frac{6}{9}\left\{n-\frac{1}{10} \frac{\left(1-\left(\frac{1}{10}\right)^{n}\right)}{\left(1-\frac{1}{10}\right)}\right\}$

$=\frac{6}{9}\left\{n-\frac{1}{9}\left(1-\frac{1}{10^{n}}\right)\right\}$