Question

Find the sum of 8+88+888+8888+...up to nth terms in a G. P.

Answer 1

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,

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