what is the sum of 5+55+555............ till 100 digits

Question

the_fizicx99 · Answer

That is alot of 5's lol, I'd say close to 1M if not more.

agent0smith · Answer

I've no idea... the sum will be huge, though... the 100th term will be a hundred 5's:

5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555

alekos · Answer

where did you get this one from?

asnaseer · Answer

$$\begin{align}
5+55+555+...&=5(1+11+111+...)\
&=5(\frac{10-1}{9}+\frac{10^2-1}{9}+...+\frac{10^{100}-1}{9})\
&=\frac{5}{9}((10-1)+(10^2-1)+...+(10^{100}-1))\
&=\frac{5}{9}((10+10^2+...+10^{100})-100)\
&=\frac{5}{9}(\frac{10(10^{100}-1)}{10-1}-100)\qquad\text{(sum of geometric series)}\
&=\frac{5}{9}(\frac{10(10^{100}-1)}{9}-100)
\end{align}$$

agent0smith · Answer

@asnaseer how did you notice the second step?

asnaseer · Answer

I noticed that (for example):$$111=1+10+100$$and then used the sum of a geometric series to obtain:$$111=1+10+100=\frac{1(10^2-1)}{10-1}=\frac{10^2-1}{9}$$

agent0smith · Answer

Very nice. I could see how you factored out the 5, but the second step... I see how you got it now, since each term is the sum of a geometric series to that term.

I used the same method for a question posted yesterday (but it was easier to spot than this one!), with this sequence: 7, 7.5, 7.75, ... since each term is $$\Large 6 + \frac{ 1-0.5^n }{ 1-0.5 }$$