| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Proposed by Paul Yiu, Florida Atlantic University, Boca Raton.
Find the smallest positive integer n so that
(n+1)^2000 > (2n+1)^1999.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1872.2 | one answer | BIGQ::DCLARK | pressure got the drop on you | Wed May 25 1994 15:36 | 13 |
re .0
(n+1)^2000 > (2n+1)^1999 =>
(n+1) > ((2n+1)/(n+1))^1999 =>
(n+1) > (2 - (1/n+1))^1999
for large n the right hand side of the inequality approaches 2, so
(n+1) > 2^1999 and the smallest integer satisfying this relationship
is n = 2^1999
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| 1872.3 | another answer | 3D::ROTH | Geometry is the real life! | Wed May 25 1994 19:44 | 1 |
n = 2^1999 - 1000 | |||||
| 1872.4 | RUSURE::EDP | Always mount a scratch monkey. | Tue Jun 13 1995 15:01 | 12 | |
Solution by Christopher J. Bradley, Clifton College, Bristol, U.K.
Expansion by the binomial theorem and division by n^1999 gives
n + 2000 + 2000*1999/n + ... > 2^1999 + 2^1998*1999/n +
2^1997(1999*1998)/(2n^2) + ... .
Since n is going to be of the order 2^1999 the terms other than the
first two on either side cannot affect the issue, leaving n to be the
first integer greater than 2^1999 + 999.5 - 2000, that is,
n = 2^1999 - 1000.
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