2007 UNCO Math Contest II Problems/Problem 7: Difference between revisions
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== Solution == | == Solution == | ||
(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>3/2</math>. | |||
(b) <math>\frac{5}{19}</math> | |||
<cmath>5T=1+\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{5}{5^4}+\cdots</cmath> | |||
<cmath>T=0+\frac{1}{5}+\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^3}+\cdots</cmath> | |||
{{ | <cmath>5T-T=1+0+\frac{1}{5^2}+\frac{1}{5^3}+\frac{2}{5^4}+\cdots = 1+\frac{T}{5}</cmath> | ||
== See Also == | == See Also == | ||
Latest revision as of 04:26, 12 January 2019
Problem
(a) Express the infinite sum 
as a reduced fraction.
(b) Express the infinite sum 
as a reduced fraction. Here the denominators are powers of 
and the numerators 
are the Fibonacci numbers
where 
.
Solution
(a): Knowing that the formula for an infinite geometric series is 
, where 
and 
are the first term and common ratio respectively, we compute 
, and we have our answer of 
.
(b) 
![\[5T=1+\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{5}{5^4}+\cdots\]](http://latex.artofproblemsolving.com/b/b/4/bb4aef8d8ac32bc4443aba716d3823d039b3e9e3.png)
![\[T=0+\frac{1}{5}+\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^3}+\cdots\]](http://latex.artofproblemsolving.com/9/3/a/93a1fae395ce571352ba2172d2a0f1212a5ca0c3.png)
![\[5T-T=1+0+\frac{1}{5^2}+\frac{1}{5^3}+\frac{2}{5^4}+\cdots = 1+\frac{T}{5}\]](http://latex.artofproblemsolving.com/8/6/1/8614c29f25462893c117b0702b73b25b7b3b266f.png)
See Also
| 2007 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| 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 | ||
