Sum of Entries in Lesser Diagonal of Pascal's Triangle equal Fibonacci Number

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Theorem

The sum of the entries in the $n$th lesser diagonal of Pascal's triangle equals the $n + 1$th Fibonacci number.


Proof

By definition, the entries in the $n$th lesser diagonal of Pascal's triangle are:

$\dbinom n 0, \dbinom {n - 1} 1, \dbinom {n - 2} 2, \dbinom {n - 3} 3, \ldots$

and so the statement can be written:

$F_{n + 1} = \ds \sum_{k \mathop \ge 0} \dbinom {n - k} k$


The proof proceeds by strong induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

$F_{n + 1} = \ds \sum_{k \mathop \ge 0} \dbinom {n - k} k$


$\map P 0$ is the case:

\(\ds \sum_{k \mathop \ge 0} \dbinom {0 - k} k\) \(=\) \(\ds \dbinom 0 0\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds F_1\)

Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

\(\ds \sum_{k \mathop \ge 0} \dbinom {1 - k} k\) \(=\) \(\ds \dbinom 1 0 + \dbinom 0 1\)
\(\ds \) \(=\) \(\ds 1 + 0\)
\(\ds \) \(=\) \(\ds F_2\)

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $0 \le j \le r$, then it logically follows that $\map P {r + 1}$ is true.


So this is the induction hypothesis:

\(\ds F_r\) \(=\) \(\ds \sum_{k \mathop \ge 0} \dbinom {r - 1 - k} k\)
\(\ds F_{r + 1}\) \(=\) \(\ds \sum_{k \mathop \ge 0} \dbinom {r - k} k\)


from which it is to be shown that:

$F_{r + 2} = \ds \sum_{k \mathop \ge 0} \dbinom {r + 1 - k} k$


Induction Step

This is the induction step:


\(\ds F_{r + 2}\) \(=\) \(\ds F_r + F_{r + 1}\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \dbinom {r - 1 - k} k + \sum_{k \mathop \ge 0} \dbinom {r - k} k\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \dbinom {r - 1} 0 + \dbinom {r - 2} 1 + \dbinom {r - 3} 2 + \dbinom {r - 4} 3 + \ldots\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \dbinom r 0 + \dbinom {r - 1} 1 + \dbinom {r - 2} 2 + \dbinom {r - 3} 3 + \dbinom {r - 4} 4 + \ldots\)
\(\ds \) \(=\) \(\ds \dbinom r 0 + \dbinom r 1 + \dbinom {r - 1} 2 + \dbinom {r - 2} 3 + \dbinom {r - 3} 4 + \ldots\) Pascal's Rule
\(\ds \) \(=\) \(\ds \dbinom {r + 1} 0 + \dbinom r 1 + \dbinom {r - 1} 2 + \dbinom {r - 2} 3 + \dbinom {r - 3} 4 + \ldots\) Binomial Coefficient with Zero
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 0} \dbinom {r + 1 - k} k\)

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{>0}: F_{n + 1} = \ds \sum_{k \mathop \ge 0} \dbinom {n - k} k$

$\blacksquare$


Sources

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