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 + 9999 + ... 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) 5 + 55 + 555 + ... to n terms

Taking 5 common form each term

5[1 + 11 + 111 + ....n~terms]

Dividing and multiplying by 9

$=\frac{5}{9}[9+99+999+...nterms]$

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

$=\frac{5}{9}$

$\left[\right\{10+{10}^2+{10}^3+...nterms\}-n]$ this is G.P.

So,

$S_n=\frac{a(r^n-1)}{r-1}-1$

$a=10,r=10,n=n$

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

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

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

(ii) 7 + 77 + 777 + .... to n terms.

Now we have

7 + 77 + 777 + ... to n terms = 7[1+ 11 + 111 + ... to n terms]

$=\frac{7}{9}[9+99+999+...$ to n terms]

$=\frac{7}{9}\left[\right(10-1)+({10}^2-1)+({10}^3-1)+....nterms]$

$=\frac{7}{9}[10+{10}^2+{10}^3+....nterms]-\frac{7}{9}(1+1+1+...tpnterms)$

$=\frac{7}{9}.\frac{10({10}^n-1)}{10-1}-\frac{7n}{9}$

$=\frac{7}{81}({10}^{n+1}-9n-10)$

(iii) 9 + 99 + 999 + .... n terms

This can be written as $=(10-1)+(100-1)+(1000-1)+...n$ terms

$=(10+{10}^2+{10}^3+....nterms)-n$

$\Rightarrow S_n=\frac{a(r^n-1)}{r-1}$, a = 10, r = 10, n = n

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

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

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

$=\frac{1}{9}[{10}^{n+1}-9n-10]$

$=9+99+999+...nterm$

(iv) $0.5+0.55+0.555+and....$ to n

$=5\times 0.1+5\times 0.11+5\times 0.11+....$

$=\frac{5}{9}\{\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+...+n\}$

$=\frac{5}{9}\{(1-\frac{1}{10})+(1-\frac{1}{100})+...+n\}$

$=\frac{5}{9}\{n-(\frac{1}{10}+\frac{1}{{10}^2}+....+\frac{1}{{10}^n})\}$

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

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

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

$=6\times 0.1+6\times 0.11+6\times 0.111+....$

$=\frac{6}{9}\{(1-\frac{1}{10})+(1-\frac{1}{100}+...+n)\}$

$=\frac{6}{9}\{n-(\frac{1}{10}+\frac{1}{{10}^2}+...+\frac{1}{{10}^n})\}$

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