Difference between revisions of "1975 USAMO Problems/Problem 3"
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− | <cmath>P(n+1) = \sum_{k=0}^n \frac{k}{k+1} \prod_{j \ | + | <cmath>P(n+1) = \sum_{k=0}^n \frac{k}{k+1} \prod_{j \ne k} \frac{n+1-j}{k-j}</cmath> |
<cmath>= \sum_{k=0}^n \frac{k}{k+1} \cdot \frac{\frac{(n+1)!}{n+1-k}}{k(k-1)(k-2) \dots 1\cdot (-1)(-2) \dots (k-n)}</cmath> | <cmath>= \sum_{k=0}^n \frac{k}{k+1} \cdot \frac{\frac{(n+1)!}{n+1-k}}{k(k-1)(k-2) \dots 1\cdot (-1)(-2) \dots (k-n)}</cmath> | ||
<cmath>= \sum_{k=0}^n \frac{k}{k+1} (-1)^{n-k}\cdot \frac{(n+1)!}{k!(n+1-k)!} </cmath> | <cmath>= \sum_{k=0}^n \frac{k}{k+1} (-1)^{n-k}\cdot \frac{(n+1)!}{k!(n+1-k)!} </cmath> |
Revision as of 18:26, 15 April 2018
Contents
Problem
If denotes a polynomial of degree
such that
for
, determine
.
Solution
Let . Clearly,
has a degree of
.
Then, for ,
.
Thus, are the roots of
.
Since these are all of the roots, we can write
as:
where
is a constant.
Thus,
Plugging in gives:
Finally, plugging in gives:
If is even, this simplifies to
. If
is odd, this simplifies to
.
Solution 2
It is fairly natural to use Lagrange's Interpolation Formula on this problem:
through usage of the Binomial Theorem.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1975 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.