Question

Find the sum to n terms of the sequences 8, 88, 888, 8888 ....

Answer 1

$S_n$ = 8 + 88 + 888 + 8888 + ....... upto n terms]

= 8 [ 1 + 11 + 111 + 1111+ ... upto n terms]

=$\frac{8}{9}$ [ 9 + 99 + 999 + 9999 + ........ upto n terms]

= $\frac{8}{9}\left[\right(10-1)+({10}^2-1)+({10}^3-1)+......uptonterms]$

= $\frac{8}{9}\left[\right(10+{10}^2+{10}^3+....nterms)-(1+1+.....nterms\left)\right]$

= $\frac{8}{9}[\frac{10.({10}^n-1)}{(10-1)}-n]$

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

= $\frac{80}{81}[{10}^n-1]-\frac{8}{9}n$