Question

Find the sum of the first n terms of 8+88+888+8888+.......

Answer 1

8 + 88 + 888 + 8888 +........up to n times
= 8(1) + 8(11) + 8(111) + 8(1111) +........up to n times
= 8{ 1 + 11 + 111 + 1111 + ....... up to n times}
= 8/9{9 + 99 + 999 + 9999 + ..... up to n times}
= 8/9 { (10-1 )+ (100-1) + (1000-1) + (10000-1) + ........up to n times}
= 8/9 { (10 + 100 + 1000 + 10000 +...... n times) - (n×1)}
= 8/9{ ( (10×(10^n -1))/(10-1) ) -n }
$8 + 88 + 888 + 8888 + ........ \\ = \frac{8}{9}( \frac{10 \times ( {10}^{n } - 1)}{10 - 1} ) \\ = \frac{80}{81} ( {10}^{n} - 1)$

Answer 2

Answer:

$\boxed{\sf \: 8+ 88 + 888 + 8888 + .. \: n \: terms = \: \dfrac{8}{9}\left[\dfrac{10( {10}^{n} - 1)}{9} - n \right] \: }$

Step-by-step explanation:

Given series is

$\sf \: 8 + 88 + 888 + 8888 + ... \: n \: terms$

$\sf \: = \: 8(1 + 11 + 111 + 1111 + ... \: n \: terms)$

$\sf \: = \: \dfrac{8}{9} (9 + 99 + 999 + 9999 + ... \: n \: terms)$

$\sf \: = \: \dfrac{8}{9} [(10 - 1) + (100 - 1) + (1000 - 1) + ... \: n \: terms]$

$\sf \: = \: \dfrac{8}{9}[ 10 + 100 + 1000 + .. \: n \: terms) - (1 + 1 + 1 + .. \: n \: terms)]$

We know,

Sum of n terms of a GP series having first term a and common ratio r is given by

$\begin{gathered}\begin{gathered}\bf\: S_n = \begin{cases} &\sf{\dfrac{a( {r}^{n} - 1) }{r - 1} }, \: \: r \: \ne \: 1 \\ \\ &\sf{\qquad \: na, \: \: \: \: r = 1} \end{cases}\end{gathered}\end{gathered}$

So, using these results, we get

$\sf \: = \: \dfrac{8}{9}\left[\dfrac{10( {10}^{n} - 1)}{10 - 1} - n \times 1 \right]$

$\sf \: = \: \dfrac{8}{9}\left[\dfrac{10( {10}^{n} - 1)}{9} - n \right]$

Hence,

$\implies\sf \: 8 + 88 + 888 + 8888 + .. \: n \: terms = \: \dfrac{8}{9}\left[\dfrac{10( {10}^{n} - 1)}{9} - n \right]$