Let a sequence of real numbers $a_0, a_1,a_2, \cdots$ satisfies the condition: $$\sum_{n=0}^ma_n\cdot(-1)^n\cdot{m\choose n}=0$$ for all sufficiently large values of $m$. Show that there exists a polynomial $P$ such that $a_n=P(n)$ for all $n\geq 0$
Problem
Source: HKTST2017
Tags: combinatorics, algebra, polynomial
guangzhou-2015
05.05.2017 07:58
This is 2016 Turkey Team Selection Test
PROF65
05.05.2017 13:44
So $\exists N / \forall m\ge N:\sum_{n=0}^ma_n\cdot(-1)^n\cdot{m\choose n}=0$ .consider a polynomial $P /$ $ \forall \ 0\le i\le N : P(i)=a_i$ we 'll proceed by induction the first step is obvious . assume that $P(n)=a_n$ we 'll prove that $P(n+1)=a_{n+1} $ : for $ n\le N-1$ it's clear so et assume $n\ge N$ we know (look at https://artofproblemsolving.com/community/c6h1441122 ) that $\sum_{k=0}^{n+1}P(k)\cdot(-1)^k\cdot{n+1\choose k}=0$ but $\sum_{k=0}^{n+1}a_k\cdot(-1)^k\cdot {n+1\choose k}=0 \implies P(n+1)=a_{n+1}$ RH HAS
Medjl
05.05.2017 15:35
i am really suprised maybe that is the first time that Hong Kong copied problems from other countries
YanYau
05.05.2017 15:37
Medjl wrote: i am really suprised maybe that is the first time that Hong Kong copied problems from other countries Definitely not, Test 3 and Test 4 last year was also copied from other countries. Generally it seems like Test 1, Test 2, and CHKMO are originally written questions, but Test 3 and Test 4 are copied.