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Mathematics-Online course: Prepcourse Mathematics - Basics - Combinatorics

Binomial Theorem

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The binomial identity provides explicit expressions for integral powers of a sum of two variables:
$\displaystyle (a+b)^n$ $\displaystyle =$ $\displaystyle a^n + \left( \begin{array}{c} n \\ 1 \end{array}\right) a^{n-1}b ... ...^2 + \cdots + \left( \begin{array}{c} n \\ n-1 \end{array}\right)ab^{n-1} + b^n$  
  $\displaystyle =$ $\displaystyle \sum_{k=0}^n \left( \begin{array}{c} n \\ k \end{array}\right)a^{n-k}b^k \,,$  

for all $ n\in\mathbb{N}_0$.

In particular, for $ n=2,3$, the formula yields

$\displaystyle (a+b)^2$ $\displaystyle =$ $\displaystyle a^2+2ab+b^2\,,$  
$\displaystyle (a+b)^3$ $\displaystyle =$ $\displaystyle a^3+3a^2 b+3ab^2 + b^3\,.$  

(Authors: Kimmerle/Abele)

The binomial theorem can be proved via mathematical induction.

For $ n = 0$ and $ n = 1$ the equation holds because of

$\displaystyle \sum_{k=0}^0 \left( \begin{array}{c} 0 \\ k \end{array}\right)a^{... ...d \sum_{k=0}^1 \left( \begin{array}{c} 1 \\ k \end{array}\right)a^{1-k}b^k=a+b $

for any arbitrarily chosen $ a$ and $ b$ .

Let us now assume that the equation holds for $ n$. This yields

$\displaystyle (a+b)^{n+1}$ $\displaystyle =$ $\displaystyle (a+b) \, (a+b)^n = (a+b) \,\sum_{k=0}^n \left( \begin{array}{c} n \\ k \end{array}\right) a^{n-k}b^k$  
  $\displaystyle =$ $\displaystyle \sum_{k=0}^n \left( \begin{array}{c} n \\ k \end{array}\right) a^... ...+ \sum_{k=0}^n \left( \begin{array}{c} n \\ k \end{array}\right) a^{n-k}b^{k+1}$  
  $\displaystyle =$ $\displaystyle a^{n+1} + \sum_{k=1}^n \left( \begin{array}{c} n \\ k \end{array}... ...n-1} \left( \begin{array}{c} n \\ k \end{array}\right) a^{n-k}b^{k+1} + b^{n+1}$  
  $\displaystyle =$ $\displaystyle a^{n+1} + \sum_{k=1}^n \left( \begin{array}{c} n \\ k \end{array}... ...1}^n \left( \begin{array}{c} n \\ k-1 \end{array}\right) a^{n-k+1}b^k + b^{n+1}$  
  $\displaystyle =$ $\displaystyle a^{n+1} + \sum_{k=1}^n \left[ \left( \begin{array}{c} n \\ k \end... ...eft( \begin{array}{c} n \\ k-1 \end{array}\right)\right] a^{n+1-k}b^k + b^{n+1}$  
  $\displaystyle =$ $\displaystyle \sum_{k=0}^{n+1} \left( \begin{array}{c} n+1 \\ k \end{array}\right) a^{n+1-k}b^k$  

(Authors: Kimmerle/Abele)
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  automatically generated 9/18/2007