I have a question about finding the sum formula of n-th terms.

Here’s the series:

5+55+555+5555+……

What is the general formula to find the sum of n-th terms?

My attempts:

I think I need to separate 5 from this series such that:

5(1+11+111+1111+....)

Then, I think I need to make the statement in the parentheses into a easier sum:

5(1+(10+1)+(100+10+1)+(1000+100+10+1)+.....)

= 5(1∗n+10∗(n−1)+100∗(n−2)+1000∗(n−3)+....)

Until the last statement, I don’t know how to go further. Is there any ideas to find the general solution from this series?

Thanks

**Answer**

5+55+555+5555+⋯+n fives⏞55…5

=59(9+99+999+9999+⋯+n nines⏞99…9)

=59(101−1+102−1+103−1+⋯+10n−1)

=59(101+102+103+⋯+10n−n)

=59(10n+1−109−n).

**Attribution***Source : Link , Question Author : akusaja , Answer Author : bof*