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Q:
Pascal's triangle and a finite sum
Let $(a_n)$ be a sequence defined as follows
$a_0=1, \qquad a_n=a_{n-1}+a_{n-2}, \quad \text{for all $n \ge 2$}.$
Let $(b_n)$ be the sequence defined by $b_0=1, b_n=a_n+b_{n-2}$ for all $n \ge 2$.
We say that $b_n$ is Pascal's triangle.
The question:
Is it true that for all $n\ge 0$,
$b_n=\sum_{j=0}^n\binom{n}{j}$?
A:
This is true. Let $c_n=b_n-\sum_{k=0}^{n-1} \binom{n}{k}$. Then $c_0=1$, $c_1=1$, $c_2=1$, and $c_n=0$ for $n\ge3$. Now, we have the recurrence
$
c_n=c_{n-1}+c_{n-2}=b_{n-1}+b_{n-2}-\binom{n-1}{0}-\binom{n-1}{1}-\cdots-\binom{n-1}{n-2}=\sum_{k=0}^{n-2} \binom{n-1}{k}.
$
Since $c_n$ is a finite 0b46394aab
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