Difference between revisions of "1969 AHSME Problems/Problem 16"
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<cmath>k^{n-2}b^{n}n(-2k+n+1)=0</cmath> | <cmath>k^{n-2}b^{n}n(-2k+n+1)=0</cmath> | ||
Dividing by <math>k^{n-2}b^{n}n</math> gives | Dividing by <math>k^{n-2}b^{n}n</math> gives | ||
| − | <cmath>n(-2k+n+1)=0 | + | <cmath>n(-2k+n+1)=0</cmath> |
Since it is given that <math>n\ge2</math>, <math>n</math> cannot equal 0, so we can divide by n, which gives | Since it is given that <math>n\ge2</math>, <math>n</math> cannot equal 0, so we can divide by n, which gives | ||
| − | <cmath>-2k+n+1=0 | + | <cmath>-2k+n+1=0</cmath> |
Solving for <math>n</math> gives | Solving for <math>n</math> gives | ||
| − | <cmath>n=2k-1 | + | <cmath>n=2k-1</cmath> so the answer is <math>\fbox{C}</math>. |
== See also == | == See also == | ||
Revision as of 12:14, 10 July 2015
Problem
When 
, is expanded by the binomial theorem, it is found that when 
, where 
is a positive integer, the sum of the second and third terms is zero. Then 
equals:
Solution
Since , we can write 
as 
.
Expanding, the second term is 
, and the third term is 
, so we can write the equation
![\[-k^{n-1}b^{n}{{n}\choose{1}}+k^{n-2}b^{n}{{n}\choose{2}}=0\]](http://latex.artofproblemsolving.com/b/4/c/b4ce32d058aa729992586d21038a517ae9798036.png)
Simplifying and multiplying by two to remove the denominator, we get
![\[-2k^{n-1}b^{n}n+k^{n-2}b^{n}n(n+1)=0\]](http://latex.artofproblemsolving.com/3/6/c/36c7dda6928e68a3cbbd0587e25e47cb62eb3c74.png)
Factoring, we get
![\[k^{n-2}b^{n}n(-2k+n+1)=0\]](http://latex.artofproblemsolving.com/1/2/0/120cca1b54dda6f351da252d2d72ebd1c5352c33.png)
Dividing by 
gives
![\[n(-2k+n+1)=0\]](http://latex.artofproblemsolving.com/3/8/3/383b6cd74d0a4a34a8242da85c008be6db58a60b.png)
Since it is given that 
, cannot equal 0, so we can divide by n, which gives
![\[-2k+n+1=0\]](http://latex.artofproblemsolving.com/d/2/0/d20febc84f8d9513ce3c6dc986dc48e1f6a6a14b.png)
Solving for gives
![\[n=2k-1\]](http://latex.artofproblemsolving.com/8/d/c/8dc8dcf056614916627e108f2152a5e42e2432d9.png)
so the answer is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
