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

Answer 1

(i) We have,
5 + 55 + 555+ ... n terms
Taking 5 as common:
$S_n$ = 5[1 + 11 + 111 + ... n terms]
$$\begin{gathered} =\frac{5}{9}\left(9+99+999+...n\mathrm{terms}\right) \\ =\frac{5}{9}\left\{\left(10-1\right)+\left({10}^2-1\right)+\left({10}^3-1\right)+...+\left({10}^n-1\right)\right\} \\ =\frac{5}{9}\left\{\left(10+{10}^2+{10}^3+...+{10}^n\right)\right\}-\left(1+1+1+1+...n\mathrm{times}\right) \\ =\frac{5}{9}\left\{10\times \frac{\left({10}^n-1\right)}{10-1}-n\right\} \\ =\frac{5}{9}\left\{\frac{10}{9}\left({10}^n-1\right)-n\right\} \\ =\frac{5}{81}\left\{{10}^{n+1}-9n-10\right\} \end{gathered}$$

(ii) We have,
7 + 77 + 777 + ... n terms
${\mathrm{S}}_n$= 7 [1 + 11 + 111 + ... n terms]
$$\begin{gathered} =\frac{7}{9}\left(9+99+999+...\mathrm{n}\mathrm{terms}\right) \\ =\frac{7}{9}\left\{\left(10-1\right)+\left({10}^2-1\right)+\left({10}^3-1\right)+...+\left({10}^n-1\right)\right\} \\ =\frac{7}{9}\left\{\left(10+{10}^2+{10}^3+...+{10}^n\right)\right\}-\left(1+1+1+1...n\mathrm{times}\right) \\ =\frac{7}{9}\left\{10\times \frac{\left({10}^n-1\right)}{10-1}-n\right\}=\frac{7}{9}\left\{\frac{10}{9}\left({10}^n-1\right)-n\right\} \\ =\frac{7}{81}\left\{{10}^{n+1}-9n-10\right\} \end{gathered}$$

(iii) We have,
9 + 99 + 999 + ... n terms
$$\begin{gathered} =\left(9+99+999+...+\mathrm{to}n\mathrm{terms}\right) \\ =\left\{\left(10-1\right)+\left({10}^2-1\right)+\left({10}^3-1\right)+...+\left({10}^n-1\right)\right\} \\ =\left\{\left(10+{10}^2+{10}^3+...+{10}^n\right)\right\}-\left(1+1+1+1...n\mathrm{times}\right) \\ =\left\{10\times \frac{\left({10}^n-1\right)}{10-1}-n\right\} \\ =\left\{\frac{10}{9}\left({10}^n-1\right)-n\right\} \\ =\frac{1}{9}\left\{{10}^{n+1}-9n-10\right\} \end{gathered}$$

(iv) We have,
0.5 + 0.55 + 0.555 + ... n terms
$S_n$= 5 [0.1 + 0.11+0.111 + ... n terms]
$$\begin{gathered} =\frac{5}{9}\left(0.9+0.99+0.999+...+\mathrm{to}n\mathrm{terms}\right) \\ =\frac{5}{9}\left\{\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...n\mathrm{terms}\right\} \\ =\frac{5}{9}\left\{\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)+...n\mathrm{terms}\right\} \\ =\frac{5}{9}\left\{n-\left(\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+...n\mathrm{terms}\right)\right\} \\ =\frac{5}{9}\left\{n-\frac{1}{10}\frac{\left(1-{\left({\displaystyle \frac{1}{10}}\right)}^n\right)}{\left(1-{\displaystyle \frac{1}{10}}\right)}\right\} \\ =\frac{5}{9}\left\{n-\frac{1}{9}\left(1-\frac{1}{{10}^n}\right)\right\} \end{gathered}$$

(v) We have,
0.6 + 0.66 +.666 + ... to n terms
$S_n$= 6 [0.1 + 0.11+ 0.111 + ... n terms]
$$\begin{gathered} =\frac{6}{9}\left(0.9+0.99+0.999+...n\mathrm{terms}\right) \\ =\frac{6}{9}\left\{\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...n\mathrm{terms}\right\} \\ =\frac{6}{9}\left\{\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)+...n\mathrm{terms}\right\} \\ =\frac{6}{9}\left\{n-\left(\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+...n\mathrm{terms}\right)\right\} \\ =\frac{6}{9}\left\{n-\frac{1}{10}\frac{\left(1-{\left({\displaystyle \frac{1}{10}}\right)}^n\right)}{\left(1-{\displaystyle \frac{1}{10}}\right)}\right\} \\ =\frac{6}{9}\left\{n-\frac{1}{9}\left(1-\frac{1}{{10}^n}\right)\right\} \end{gathered}$$