SOLUTION: Find the sum \frac{1}{7^" "} + \frac{2}{7^2} + \frac{3}{7^3} + \frac{1}{7^4} + \frac{2}{7^5} + \frac{3}{7^6}

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Question 1209811: Find the sum
\frac{1}{7^" "} + \frac{2}{7^2} + \frac{3}{7^3} + \frac{1}{7^4} + \frac{2}{7^5} + \frac{3}{7^6}
Found 2 solutions by CPhill, KMST:
Answer by CPhill(2303) (Show Source):
You can put this solution on YOUR website!
Let $S = \frac{1}{7} + \frac{2}{7^2} + \frac{3}{7^3} + \frac{1}{7^4} + \frac{2}{7^5} + \frac{3}{7^6}$.
We can group the terms in sets of three:
$S = \left(\frac{1}{7} + \frac{2}{7^2} + \frac{3}{7^3}\right) + \left(\frac{1}{7^4} + \frac{2}{7^5} + \frac{3}{7^6}\right)$.
Let $A = \frac{1}{7} + \frac{2}{7^2} + \frac{3}{7^3}$.
Then $S = A + \frac{1}{7^3} A = A\left(1 + \frac{1}{7^3}\right)$.
We have $A = \frac{1}{7} + \frac{2}{49} + \frac{3}{343} = \frac{49}{343} + \frac{14}{343} + \frac{3}{343} = \frac{49+14+3}{343} = \frac{66}{343}$.
Then $S = \frac{66}{343}\left(1 + \frac{1}{343}\right) = \frac{66}{343}\left(\frac{343+1}{343}\right) = \frac{66}{343}\left(\frac{344}{343}\right) = \frac{66\cdot 344}{343\cdot 343} = \frac{22704}{117649}$.
We can also write
$S = \sum_{k=0}^1 \left( \frac{1}{7^{3k+1}} + \frac{2}{7^{3k+2}} + \frac{3}{7^{3k+3}} \right) = \sum_{k=0}^1 \frac{1 \cdot 7^2 + 2 \cdot 7 + 3}{7^{3k+3}} = \sum_{k=0}^1 \frac{49+14+3}{7^{3k+3}} = \sum_{k=0}^1 \frac{66}{7^{3k+3}}$.
$S = \frac{66}{7^3} + \frac{66}{7^6} = \frac{66}{343} + \frac{66}{117649} = \frac{66\cdot 343 + 66}{117649} = \frac{66(343+1)}{117649} = \frac{66(344)}{117649} = \frac{22704}{117649}$.
Final Answer: The final answer is $\boxed{\frac{22704}{117649}}$

Answer by KMST(5484) (Show Source):
You can put this solution on YOUR website!
Translating the original question and CPhill's answer to a language we can all understand (from I think LaTex):

That is an irreducible fraction, because numerator and denominator have no prime factors in common