Difference between revisions of "1991 AHSME Problems/Problem 25"
(Created page with "== Problem == If <math>T_n=1+2+3+\cdots +n</math> and <cmath>P_n=\frac{T_2}{T_2-1}\cdot\frac{T_3}{T_3-1}\cdot\frac{T_4}{T_4-1}\cdot\cdots\cdot\frac{T_n}{T_n-1}</cmath> for <math...") |
(Added a solution with explanation) |
||
| Line 12: | Line 12: | ||
== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | <math>\fbox{D}</math> It is well known that <math>T_n = \frac{n(n+1)}{2}</math>, so we can put this in and simplify to get that the general term of the product is <math>\frac{n(n+1)}{(n+2)(n-1)}.</math> Now by writing out the terms, we see that almost everything cancels (telescopes), giving <math>\frac{3 \times 1991}{1993}</math> which is almost <math>3</math>, so closest to <math>2.9.</math> |
== See also == | == See also == | ||
Latest revision as of 02:18, 24 February 2018
Problem
If 
and
![\[P_n=\frac{T_2}{T_2-1}\cdot\frac{T_3}{T_3-1}\cdot\frac{T_4}{T_4-1}\cdot\cdots\cdot\frac{T_n}{T_n-1}\]](http://latex.artofproblemsolving.com/e/1/a/e1a851ae876f5c6fdd2ecf422cf34dea5012a5e3.png)
for 
then 
is closest to which of the following numbers?
Solution
It is well known that 
, so we can put this in and simplify to get that the general term of the product is 
Now by writing out the terms, we see that almost everything cancels (telescopes), giving 
which is almost 
, so closest to 
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
