Question

Compute the sum S_{n}=9+99+999+....+999....999 .

Answer 1

Assume the last number 999..999 has n digits, then
\(\large S_n=\sum_{k=1}^n 10^{k}-1.\)

Answer 2

That's because 9=10-1, 99=10^2-1, 999=10^3-1, .. and so on.