I have the series: 1, 13, 133, 1333 ...
Currently I have distributed it down to: 1 + (10 * 2) + (100 * 2) ... Can anyone point me in the right direction?
Sorry, I forgot to mention, I'm looking for the sum of the series up to n where n is the stage.
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Hint
If you multiply all numbers by $3$, you have $$3,39,399,3999,\cdots$$ that is to say $$4-1,40-1,400-1,4000-1,\cdots $$ which look like ... something I let you.
I am sure you can take from here.
Claude Leibovici
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Hello Claude, I am sorry but I am still not following how this can get the sum of the series. – fredthadinosaur Feb 18 '15 at 10:43
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Just use the suggested formula for $$a_n=\frac{1}{3} (4\times 10^n - 1)= \frac{4}{3} \times 10^n - \frac{4}{3}$$ and sum from $n=0$ to $n=p$. The first part is a geometric progression and the second part looks quite simple. – Claude Leibovici Feb 18 '15 at 11:08
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Sorry Again, I have found the pattern doesn't seem to work on stage 0. Am I doing something wrong? – fredthadinosaur Feb 18 '15 at 12:27
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Sorry for the typo $a_n=\frac{4}{3} \times 10^n - \frac{1}{3}$ and this works for $n=0$ – Claude Leibovici Feb 18 '15 at 17:27
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That then equals 1? shouldnt it equal 0? – fredthadinosaur Feb 18 '15 at 23:16