Binomial Theorem/Integral Index

From ProofWiki
Jump to navigation Jump to search


Theorem

Let $X$ be one of the standard number systems $\N$, $\Z$, $\Q$, $\R$ or $\C$.

Let $x, y \in X$.

Then:

\(\ds \forall n \in \Z_{\ge 0}: \, \) \(\ds \paren {x + y}^n\) \(=\) \(\ds \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k\)
\(\ds \) \(=\) \(\ds x^n + \binom n 1 x^{n - 1} y + \binom n 2 x^{n - 2} y^2 + \binom n 3 x^{n - 3} y^3 + \cdots\)
\(\ds \) \(=\) \(\ds x^n + n x^{n - 1} y + \frac {n \paren {n - 1} } {2!} x^{n - 2} y^2 + \frac {n \paren {n - 1} \paren {n - 3} } {3!} x^{n - 3} y^3 + \cdots\)

where $\dbinom n k$ is $n$ choose $k$.


Proof

Basis for the Induction

For $n = 0$ we have:

$\ds \paren {x + y}^0 = 1 = \binom 0 0 x^{0 - 0} y^0 = \sum_{k \mathop = 0}^0 \binom 0 k x^{0 - k} y^k$

This is the basis for the induction.


Induction Hypothesis

This is our induction hypothesis:

$\ds \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$


Induction Step

This is our induction step:

\(\ds \paren {x + y}^{n + 1}\) \(=\) \(\ds \paren {x + y} \paren {x + y}^n\)
\(\ds \) \(=\) \(\ds x \sum_{k \mathop = 0}^n \binom n k x^{n - k}y^k + y \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k\) Inductive Hypothesis
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \binom n k x^{n + 1 - k} y^k + \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^{k + 1}\)
\(\ds \) \(=\) \(\ds \binom n 0 x^{n + 1} + \sum_{k \mathop = 1}^n \binom n k x^{n + 1 - k} y^k + \binom n n y^{n + 1} + \sum_{k \mathop = 0}^{n - 1} \binom n k x^{n - k} y^{k + 1}\)
\(\ds \) \(=\) \(\ds x^{n + 1} + y^{n + 1} + \sum_{k \mathop = 1}^n \binom n k x^{n + 1 - k} y^k + \sum_{k \mathop = 0}^{n - 1} \binom n k x^{n - k} y^{k + 1}\)
\(\ds \) \(=\) \(\ds \binom {n + 1} 0 x^{n + 1} + \binom {n + 1} {n + 1} y^{n + 1} + \sum_{k \mathop = 1}^n \binom n k x^{n + 1 - k} y^k + \sum_{k \mathop = 1}^n \binom n {k - 1} x^{n + 1 - k} y^k\)
\(\ds \) \(=\) \(\ds \binom {n + 1} 0 x^{n + 1} + \binom {n + 1} {n + 1} y^{n + 1} + \sum_{k \mathop = 1}^n \paren {\binom n k + \binom n {k - 1} } x^{n + 1 - k} y^k\)
\(\ds \) \(=\) \(\ds \binom {n + 1} 0 x^{n + 1} + \binom {n + 1} {n + 1} y^{n + 1} + \sum_{k \mathop = 1}^n \binom {n + 1} k x^{n + 1 - k} y^k\) Pascal's Rule
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{n + 1} \binom {n + 1} k x^{n + 1 - k} y^k\)

The result follows by the Principle of Mathematical Induction.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Binomial_Theorem/Integral_Index&oldid=626763"