Higher-Order Arithmetic Sequences and Differences

What this move does

Take differences of neighboring terms of a sequence $\{a_n\}$ and you get the first differences $\Delta a_n=a_{n+1}-a_n$. Take differences of those and you get the second differences ${\Delta }^2{a}_n$, and so on. If the $p$-th differences are a non-zero constant, $\{a_n\}$ is called an arithmetic sequence of order $p$. (An arithmetic sequence of order $1$ is just an ordinary arithmetic sequence.)

The key fact: the general term of an arithmetic sequence of order $p$ is exactly a polynomial of degree $p$ in $n$. In $(n+1{)}^p-n^p$ the leading terms cancel, so each time you take differences, the degree of the polynomial drops by one.

1. Take differences row by row until a constant row appears.
2. Write one more copy of the constant at the end of that row, and add back up: this gives the next term.
3. Use the left-hand column of the table to write down the general term (Newton's formula):

Work through one

Why is the left-hand column enough? ${\mathrm{C}}_{n-1}^k$ is $0$ for $n=1,2,\dots ,k$ and equals $1$ at $n=k+1$. So at $n=1$ Newton's formula leaves only $a_1$; at $n=2$ it leaves only $a_1+\Delta a_1=a_2$; and so on. It is a formula that “locks onto one term at a time”, which matches the structure of the difference table exactly.

When it fails

Practice