Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 7"

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== Solution ==  
 
== Solution ==  
  
Part A: Knowing that the formula for an infinite geometric series is <math>A/(1 - r)</math>, where <math>A</math> and <math>r</math> are the first term and common ratio respectively, we compute <math>1/(1 - 1/3) = 3/2</math>, and therefore, we have our answer of <math>2/3</math>.  
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Part A: Knowing that the formula for an infinite geometric series is <math>A/(1 - r)</math>, where <math>A</math> and <math>r</math> are the first term and common ratio respectively, we compute <math>1/(1 - 1/3) = 3/2</math>, and we have our answer of <math>2/3</math>.  
  
  

Revision as of 21:05, 5 December 2016

Problem

(a) Express the infinite sum $S= 1+ \frac{1}{3}+\frac{1}{3^2}+ \frac{1}{3^3}+ \cdots$ as a reduced fraction.

(b) Express the infinite sum $T=\frac{1}{5}+ \frac{1}{25}+ \frac{2}{125}+ \frac{3}{625}+ \frac{5}{3125}+ \cdots$ as a reduced fraction. Here the denominators are powers of $5$ and the numerators $1, 1, 2, 3, 5, \ldots$ are the Fibonacci numbers $F_n$ where $F_n=F_{n-1}+F_{n-2}$.

Solution

Part A: Knowing that the formula for an infinite geometric series is $A/(1 - r)$, where $A$ and $r$ are the first term and common ratio respectively, we compute $1/(1 - 1/3) = 3/2$, and we have our answer of $2/3$.


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See Also

2007 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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