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).
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