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
Best Answer
$$5+55+555+5555+\cdots+\overbrace{55\dots5}^{n\text{ fives}}$$ $$=\frac59(9+99+999+9999+\cdots+\overbrace{99\dots9}^{n\text{ nines}})$$ $$=\frac59(10^1-1+10^2-1+10^3-1+\cdots+10^n-1)$$ $$=\frac59(10^1+10^2+10^3+\cdots+10^n-n)$$ $$=\frac59\left(\frac{10^{n+1}-10}{9}-n\right).$$