Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?

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).


Statistics:

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