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The difference table for the sequence $a_0, a_1, a_2, a_3,\cdots$ is
sequence $a_0$ $a_1$ $a_2$ $a_3$ $a_4$ $a_5$ $a_6$ ...
first difference $b_0$ $b_1$ $b_2$ $b_3$ $b_4$ $b_5$ ...
second difference $c_0$ $c_1$ $c_2$ $c_3$ $c_4$ ...
third difference 0 0 0 0 ...

Show that \begin{eqnarray} a_n &=& a_0\left( \begin{array}{r}n\\0\end{array} \right) + b_0\left( \begin{array}{r}n\\1\end{array} \right)+c_0\left( \begin{array}{r}n\\2\end{array} \right). \end{eqnarray}

J. W. Tanner
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Paul Steiner
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1 Answers1

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The fourth row says that for all $n\ge0$, $c_{n+1}-c_n=0$, so $c_n=c_0$ is a constant sequence.

Then $b_n$ is a linear sequence, with

$$c_n=b_{n+1}-b_n \implies b_{n+1}=b_n+c_0 \implies b_n=b_0+c_0n$$

and $a_n$ is a quadratic sequence, with

$$b_n=a_{n+1}-a_n \implies a_{n+1} = a_n+b_0+c_0n \implies a_n = a_0 + b_0n + c_0\sum_{k=1}^n(n-k)$$

Now,

$$\sum_{k=1}^n(n-k)=n^2-\frac{n(n+1)}2=\frac{n(n-1)}2=\binom n2$$

and of course $\binom n0=1$ and $\binom n1=n$.

user170231
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