Rising Sum of Binomial Coefficients/Proof by Induction
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Theorem
- $\ds \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1} = \binom {n + m + 1} m$
Proof
Proof by induction:
Let $n \in \Z$.
For all $m \in \N$, let $\map P m$ be the proposition:
- $\ds \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$
$\map P 0$ is true, as this just says:
- $\dbinom n n = \dbinom {n + 1} {n + 1}$
But $\dbinom n n = \dbinom {n + 1} {n + 1} = 1$ from the Definition of Binomial Coefficient.
Basis for the Induction
$\map P 1$ is the case:
\(\ds \sum_{j \mathop = 0}^1 \binom {n + j} n\) | \(=\) | \(\ds \binom n n + \binom {n + 1} n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \paren {n + 1}\) | Definition of Binomial Coefficient | |||||||||||
\(\ds \) | \(=\) | \(\ds n + 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom {n + 2} {n + 1}\) | Definition of Binomial Coefficient |
So:
- $\ds \sum_{j \mathop = 0}^1 \binom {n + j} n = \binom {n + 2} {n + 1}$
and $\map P 1$ is seen to hold.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\ds \sum_{j \mathop = 0}^k \binom {n + j} n = \binom {n + k + 1} {n + 1}$
Then we need to show:
- $\ds \sum_{j \mathop = 0}^{k+1} \binom {n + j} n = \binom {n + k + 2} {n + 1}$
Induction Step
This is our induction step:
\(\ds \sum_{j \mathop = 0}^{k + 1} \binom {n + j} n\) | \(=\) | \(\ds \sum_{j \mathop = 0}^k \binom {n + j} n + \binom {n + k + 1} n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom {n + k + 1} {n + 1} + \binom {n + k + 1} n\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {n + k + 2} {n + 1}\) | Pascal's Rule |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$
Finally, from Symmetry Rule for Binomial Coefficients:
- $\dbinom {n + m + 1} {n + 1} = \dbinom {n + m + 1} m$
$\blacksquare$
Proof
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {3-1}$ Permutations and Combinations: Exercise $5$