Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 25, 2011"
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{{:AoPSWiki:Problem of the Day/June 24, 2011}} | {{:AoPSWiki:Problem of the Day/June 24, 2011}} | ||
==Solutions== | ==Solutions== | ||
| − | {{ | + | The sum in general form is |
| + | <math>\sum_{i=1}^{\infty}\frac{1}{(3n-1)(3n+2)}</math> | ||
| + | We wish to write <math>\frac{1}{(3n-1)(3n+2)}</math> as <math>\frac{A}{3n-1} + \frac{B}{3n+2}</math> | ||
| + | We that by setting them equal that <math>n(3A + 3B) = 0</math> and <math>2A-B=1</math>. | ||
| + | We find that <math>A=\frac{1}{3}</math> and <math>B=\frac{-1}{3}</math> | ||
| + | We now note that <math>\frac{\frac{1}{3}}{3n-1} - \frac{\frac{1}{3}}{3n+2} = \frac{1}{(3n-1)(3n+2)}</math> | ||
| + | We now rewrite the sum as <math>\sum_{i=1}^{\infty}\frac{\frac{1}{3}}{3n-1} - \frac{\frac{1}{3}}{3n+2}</math> | ||
| + | Now we note that <math>3(n+1)-1=3n+3-1=3n+2</math>. As a result, this is a telescoping sum. Hence, the total sum is the first term or <math>\boxed{\frac{1}{6}}</math> | ||
Latest revision as of 18:25, 25 June 2011
Problem
AoPSWiki:Problem of the Day/June 24, 2011
Solutions
The sum in general form is
We wish to write 
as 
We that by setting them equal that 
and 
.
We find that 
and 
We now note that 
We now rewrite the sum as 
Now we note that 
. As a result, this is a telescoping sum. Hence, the total sum is the first term or 
