Find the sum of 1 + 11 + 111 + 1111 + ...

[Math Tutor]

What is the the formula for the sum of numbers [tex]1 + 11 + 111 + 1111 + ... + 11111_{(n)}[/tex]

[henrik_sar]

[10][111.1_n][-n][/9]

[Math Tutor]

I do not understand your solution of this problem of the month.

[Henrik]

Hi Dear teacher
using the notification here is very hard to understand
so I write the answer like this:
(10 times the n-th term minus n)/9

[Math Tutor]

The answer of the problem is wrong!
You should think a little bit more.

[userfool]

OK i give the steps only because i am not skillful of writing formulas here.
Any member of this sum is a sum of a geometric progression. So a_n = 10^0 + 10^1 + + 10^(n-1). By the formula for the sum of a geometric progression we have a_n = (10^n - 1)/9

Then we have S = a_1 + a_2 + .... = 1/9(10^1 + 10^2 + ... + 10^n - n)
After some simle manipulationn the result is : 1/81(10^(n+1) - 9n - 10)

[Math Tutor]

Yes, the result is correct.
The sum of 1 + 11 + 111 +...+ 1111(n) = 1/81(10^(n+1) - 9n - 10)