Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 14, 2011"
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{{:AoPSWiki:Problem of the Day/July 14, 2011}} | {{:AoPSWiki:Problem of the Day/July 14, 2011}} | ||
==Solution== | ==Solution== | ||
| − | {{ | + | We begin by factoring the given expression, <math>\prod_{n=1}^{39}\frac{n^2+6n+9}{n^2+6n+8}</math>, to <math>\prod_{n=1}^{39}\frac{(n+3)^2}{(n+2)(n+4)}</math>. |
| + | Then, writing this as multiple products and shifting the indices for clarity, we get <math>\frac{(\prod_{n=4}^{42}n)^2}{(\prod_{n=3}^{41}n)(\prod_{n=5}^{43}n)}</math>. | ||
| + | Clearly, this equals <math>\frac{(\frac{42!}{6})^2}{(\frac{41!}{2})(\frac{43!}{24})}</math>. | ||
| + | At this point, all that is left is arithmetic. | ||
| + | The expression equals <math>\frac{(42!)^2*2*24}{41!*43!*6^2}=\frac{42!}{41!}*\frac{42!}{43!}*\frac{48}{36}=42*\frac{1}{43}*\frac{4}{3}</math>. | ||
| + | This trivially simplifies to <math>\frac{168}{129}=\boxed{\frac{56}{43}}</math>. | ||
Latest revision as of 11:58, 20 June 2012
Problem
AoPSWiki:Problem of the Day/July 14, 2011
Solution
We begin by factoring the given expression, 
, to 
.
Then, writing this as multiple products and shifting the indices for clarity, we get 
.
Clearly, this equals 
.
At this point, all that is left is arithmetic.
The expression equals 
.
This trivially simplifies to 
.
