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 Algebra Pre-Algebra and Basic Algebra Math Forum

May 29th, 2015, 03:29 AM   #1
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My poor brain can't handle this sequence

Hi! I am having trouble with this sequence. I found the general term An but I dont know how to find the sum. Can someone help me?
Attached Images sequence 3.jpg (25.4 KB, 9 views) May 29th, 2015, 04:04 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Let \begin{aligned} s &= \phantom{2 + } \frac23 + \frac49 + \frac6{27} + \cdots \\ &= \phantom{2 + } {2 \over 3} + {4 \over 3^2} + {6 \over 3^3} + \cdots \\ 3s &= 2 + {4 \over 3} + {6 \over 3^2} + {8 \over 3^3} + \cdots \\ 3s - s &= 2 + {2 \over 3} + {2 \over 3^2} + {2 \over 3^3} + \cdots \\ &= 2\left( 1 + \left(\frac13\right) + \left(\frac13\right)^2 + \left(\frac13\right)^3 + \cdots\right) \end{aligned} You should be able easily to evaluate the last bracket to leave yourself with a simple piece of algebra to find $s$. Thanks from Country Boy and matisolla May 29th, 2015, 05:09 AM #3 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms That's the infinite sum. The finite sum is $$S=r+r^2+\cdots+r^n$$ which satisfies $$rS=r^2+r^3+\cdots+r^{n+1}$$ so $$S-rS=r-r^{n+1}$$ and thus $$S=\frac{r-r^{n+1}}{1-r}.$$ I used "$r$" in place of "2/3" for generality. Thanks from matisolla May 29th, 2015, 10:37 AM   #4
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Hello, matisolla!

Quote:

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