Question

Sum upto $n$ terms the series

$8+88+888+.........$ be $\frac{m}{81}[k^{n+1}-9n-k]$.FInd $k-m$ ?

Answer 1

$S=8+88+888+...$

$=8+80+8+800+80+8+...$

$=8+8\left(10\right)+8\left({10}^2\right)+8\left(10\right)+\mathrm{8...}$

$S=8(n+(n-1)10+(n-2){10}^2+...{10}^{n-1})$ .....(1)

$\Rightarrow 10S=8(n10+(n-1){10}^2+,...{10}^n)$ .........(2)

Substract equation(2) from equation(1)

$\Rightarrow (1-10)S=8[n-10-{10}^2-...-{10}^{n-1}+{10}^n]$

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

$\therefore k-m=10-8=2$