\section*{Problem 1 - 091221}

Given a sequence of real numbers \(a_{1}, a_{2}, \ldots, a_{n}, \ldots\) by the (independent) representation

\[
a_{n}=c_{2} n^{2}+c_{1} n+c_{0}
\]

where \(c_{0}, c_{1}, c_{2}\) are real numbers. The first difference sequence is denoted by \(D(1)_{n}=a_{n+1}-a_{n}\) and the second difference sequence by \(D(2)_{n}=D(1)_{n+1}-D(1)_{n},(n=1,2,3, \ldots)\).

a) Let \(c_{0}=1, c_{1}=-1, c_{2}=1\). Under this assumption, \(a_{n}, D(1)_{n}, D(2)_{n}\) are to be calculated for \(n=1,2,3,4\) and 5.

b) It is to be proven generally that for (1) the sequence \(D(2)_{n}\) is constant.
}

a)

\begin{tabular}{r|r|r|r|r|r|r|r}
\(n\) & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline\(a_{n}\) & 1 & 3 & 7 & 13 & 21 & 31 & 43 \\
\hline\(D(1)_{n}\) & 2 & 4 & 6 & 8 & 10 & 12 & \\
\hline\(D(2)_{n}\) & 2 & 2 & 2 & 2 & 2 & &
\end{tabular}

b) It is for all non-negative integers \(n\)

\[
\begin{gathered}
D(1)_{n}=a_{n+1}-a_{n}=c_{2}\left((n+1)^{2}-n^{2}\right)+c_{1}((n+1)-n)+c_{0}-c_{0}=2 c_{2} n+c_{2}+c_{1} \quad \text { and } \\
D(2)_{n}=D(1)_{n+1}-D(1)_{n}=2 c_{2}((n+1)-n)+\left(c_{2}+c_{1}\right)-\left(c_{2}+c_{1}\right)=2 c_{2}
\end{gathered}
\]

which proves the claim,