Sum of Sequence of n Choose 2/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \Z$ be an integer such that $n > 2$.

\(\displaystyle \sum_{j \mathop = 2}^n \dbinom j 2\) \(=\) \(\displaystyle \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2\)
\(\displaystyle \) \(=\) \(\displaystyle \dbinom {n + 1} 3\)

where $\dbinom n j$ denotes a binomial coefficient.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:

$\displaystyle \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$


Basis for the Induction

$\map P 2$ is the case:

\(\displaystyle \sum_{j \mathop = 2}^2 \dbinom j 2\) \(=\) \(\displaystyle \dbinom 2 2\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)
\(\displaystyle \) \(=\) \(\displaystyle \dbinom 2 2\)

Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\sum_{j \mathop = 2}^k \dbinom j 2 = \dbinom {k + 1} 3$


from which it is to be shown that:

$\sum_{j \mathop = 2}^{k + 1} \dbinom j 2 = \dbinom {k + 2} 3$


Induction Step

This is the induction step:

\(\displaystyle \sum_{j \mathop = 2}^{k + 1} \dbinom j 2\) \(=\) \(\displaystyle \sum_{j \mathop = 2}^k \dbinom j 2 + \dbinom {k + 1} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \dbinom {k + 1} 3 + \dbinom {k + 1} 2\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \dbinom {k + 2} 3\) Pascal's Rule

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


Therefore:

$\forall n \in \Z_{\ge 2}: \displaystyle \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$


Sources