Question

Find the sum of $8+88+888+8888+\cdots$

Answer 1

$8+88+888+8888+....$

$=8(1+11+111+1111+...)$

$=\frac{8}{9}(9+99+999+9999+...)$

$=\frac{8}{9}(10-1+100-1+1000-1+...)$

$=\frac{8}{9}(10+{10}^2+{10}^3+...+{10}^n-n)$

$=\frac{8}{9}(\frac{10({10}^n-1)}{10-1}-n)$ [sum of n terms of a G.P]

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