Question

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

Answer 1

$8+88+888+\mathrm{8888...........}8\left[1+11+111+1111+.....\right]\frac{8}{9}\left[9+99+999+.....\right]\frac{8}{9}\left[\left(10-1\right)+\left(100-1\right)+\left(1000-1\right)+........tonterm\right]\frac{8}{9}\left[\left\{10+100+1000+.......tonterm\right\}-n\right]\frac{8}{9}\left[\left(10+{10}^2+{10}^3+.......+tonterm\right)-n\right]\frac{8}{9}\left\{\frac{10\left({10}^{n-1}\right)}{\left(10-1\right)}-n\right\}=\frac{8}{9}\left(\frac{{10}^{(n+1)}-10-9n}{9}\right)=\frac{8}{81}\left({10}^{\left(n+1\right)}-9n-10\right)Ans.$