Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem A. 379. (September 2005)

A. 379. Find all real numbers \lambda for which there exists a non-zero polynomial P, such that

\frac{P(1)+P(3)+P(5)+\dots+P(2n-1)}{n} = \lambda P(n)

for all n. List all such polynomials for \lambda=2.

(5 pont)

Deadline expired on October 17, 2005.

Sketch of solution. Let P(x)=a0+a1x+...+akxk. Both sides of the equation contains a polynomial of n, degree is k. The coefficients must pairwise match.

Since \lim\frac{1^k+3^k+\dots+(2n-1)^k}{n^{k+1}}=\frac{2^k}{k+1}, the coefficient of nk is \frac{2^k}{k+1}a_k on the left-hand side, and it is \lambdaak on the right-hand side. Therefore, \lambda=\frac{2^k}{k+1}.

If \lambda=\frac{2^k}{k+1} then we have a homogenous system of linear equations for the coefficients and one of these equations vanishes. So the system has infinitely many solutions.

For \lambda=2 we obtain k=3 and P(x)=c(x3-x).


15 students sent a solution.
5 points:Erdélyi Márton, Estélyi István, Gyenizse Gergő, Kónya 495 Gábor, Nagy 224 Csaba, Paulin Roland, Tomon István.
4 points:Jankó Zsuzsanna, Molnár 999 András.
3 points:2 students.
1 point:1 student.
0 point:3 students.

Problems in Mathematics of KöMaL, September 2005