Correct Answer - Option 4 :
\(\frac{{80}}{{81}}\left( {{{10}^n} - 1} \right) - \frac{8}{9}n\)
Concepts:
Let us consider sequence a1, a2, a3 …. an is an G.P.
- Common ratio = r = \(\frac{{{a_2}}}{{{a_1}}} = \frac{{{a_3}}}{{{a_2}}} = \ldots = \frac{{{a_n}}}{{{a_{n - 1}}}}\)
- nth term of the G.P. is an = arn−1
- Sum of n terms = s = \(\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - 1}}\); where r >1
- Sum of n terms = s = \(\frac{{a\;\left( {1 - {r^n}} \right)}}{{1 - r}}\); where r <1
- Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{}}\frac{{\rm{a}}}{{1{\rm{}} - {\rm{r}}}}{\rm{}}\); |r| < 1
Where a is 1st term and r is common ratio.
Explanation:
8 + 88 + 888 + 8888 + ….
Taking 8 common from each term
⇒ 8 (1 + 11 + 111 + 1111 + ….)
Multiplying and dividing the equation by 9
⇒ \(\frac{8}{9}{\rm{\;}}\left[ {9 + 99 + 999 + 9999{\rm{\;}} + {\rm{\;}} \ldots .} \right]\)
⇒ \(\frac{8}{9}{\rm{\;}}\left[ {\left( {10 - 1} \right) + \left( {100 - 1} \right) + \left( {1000 - 1} \right) + \left( {10000 - 1} \right) + {\rm{\;}} \ldots .} \right]\)
⇒ \(\frac{8}{9}{\rm{\;}}\left[ {\left( {10 + {{10}^2} + {{10}^3} + {{10}^4} + {\rm{\;}} \ldots + {{10}^{\rm{n}}}} \right) - \left( {1 + 1 + 1 + 1 + {\rm{\;}} \ldots .{\rm{\;}} + {\rm{\;n}}} \right)} \right]\)
The above equation is in geometric progression (GP) with a ratio of 10.
∵ the sum of G.P of with first term a, geometric ratio r and number of terms n is given by \(\frac{{{\bf{a}}\left( {{{\bf{r}}^{\bf{n}}} - 1} \right)}}{{{\bf{r}} - 1}}\)
⇒ \({\rm{\;}}\frac{8}{9}{\rm{\;}}\left[ {\left\{ {\frac{{10\left( {{{10}^{\rm{n}}} - 1} \right)}}{9}} \right\} - \left( {\rm{n}} \right)} \right]\)
⇒ \(\;\frac{{80\left( {{{10}^{\bf{n}}} - 1} \right)}}{{81}} - \frac{{8{\bf{n}}}}{9}\)
Trick:
You can also use the formula to find the sum of such problems.
\(\frac{{\rm{N}}}{9}\;\left[ {\left\{ {\frac{{10\left( {{{10}^n} - 1} \right)}}{9}} \right\} - \left( n \right)} \right]\)
Here N is the single-digit no which is repeating
For example: In this question, N = 8