Sum of Sequence of n Choose 2/Proof 2
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Theorem
Let $n \in \Z$ be an integer such that $n \ge 2$.
\(\ds \sum_{j \mathop = 2}^n \dbinom j 2\) | \(=\) | \(\ds \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom {n + 1} 3\) |
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:
- $\ds \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$
Basis for the Induction
$\map P 2$ is the case:
\(\ds \sum_{j \mathop = 2}^2 \dbinom j 2\) | \(=\) | \(\ds \dbinom 2 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom 3 3\) |
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:
- $\ds \sum_{j \mathop = 2}^k \dbinom j 2 = \dbinom {k + 1} 3$
from which it is to be shown that:
- $\ds \sum_{j \mathop = 2}^{k + 1} \dbinom j 2 = \dbinom {k + 2} 3$
Induction Step
This is the induction step:
\(\ds \sum_{j \mathop = 2}^{k + 1} \dbinom j 2\) | \(=\) | \(\ds \sum_{j \mathop = 2}^k \dbinom j 2 + \dbinom {k + 1} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom {k + 1} 3 + \dbinom {k + 1} 2\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom {k + 2} 3\) | Pascal's Rule |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{\ge 2}: \ds \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $4 \ \text {(a)}$