Intermediate Algebra/Binomial Theorem

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[edit] Algebra

The binomial thereom gives the coefficients of the polynomial

(x + y) n

.

We may consider without loss of generality the polynomial, of order n, of a single variable z. Assuming $x \ne 0$ set z = y / x

(x + y) n = x n (1 + z) n

.

The expansion coefficients of $(1 + z) n$ are known as the binomial coefficients, and are denoted

$(1 + z)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^k$ .

Noting that

$(x + y)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^{n - k} y^k$

is symmetric in x and y, the identity

$\begin{pmatrix} n \\ n - k \end{pmatrix} = \begin{pmatrix} n \\ k \end{pmatrix}$

may be shown by replacing k by n - k and reversing the order of summation.

A recursive relationship between the $\begin{pmatrix} n \\ k \end{pmatrix}$ may be established by considering

$(1 + z)^{n+1} = (1 + z)(1 + z)^n = \sum_{k=0}^{n + 1} \begin{pmatrix} n + 1 \\ k \end{pmatrix} z^k = (1 + z) \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^k$

or

$\sum_{k=0}^{n + 1} \begin{pmatrix} n + 1 \\ k \end{pmatrix} z^k = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^k + \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^{k + 1} = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} z^k + \sum_{k=1}^{n + 1} \begin{pmatrix} n \\ k - 1 \end{pmatrix} z^k$ .

Since this must hold for all values of z, the coefficients of $z k$ on both sides of the equation must be equal

$\begin{pmatrix} n + 1 \\ k \end{pmatrix} = \begin{pmatrix} n \\ k \end{pmatrix} + \begin{pmatrix} n \\ k - 1 \end{pmatrix}$

for k ranging from 1 through n, and

$\begin{pmatrix} n + 1 \\ n + 1 \end{pmatrix} = \begin{pmatrix} n \\ n \end{pmatrix} = 1$

$\begin{pmatrix} n + 1 \\ 0 \end{pmatrix} = \begin{pmatrix} n \\ 0 \end{pmatrix} = 1$ .

Pascal's Triangle is a schematic representation of the above recursion relation ...

Show

$\begin{pmatrix} n \\ k \end{pmatrix} = \frac{n!}{k!(n - k)!}$

(proof by induction on n).

A useful identity results by setting $z = 1$

$\sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} = 2^n$ .

[edit] The visual way to do the binomial theorem

Wikipedia has related information at Pascal's triangle.

(this section is from difference triangles)

Lets look at the results for (x+1)^n where n ranges from 0 to 3.

(X+1)^0 =           1X^0          =                   1         
(X+1)^1 =        1X^1+1X^0        =                 1   1
(X+1)^2 =      1X^2+2X^1+1X^0     =               1   2   1
(X+1)^3 =    1X^3+3X^2+3X^1+1X^0  =             1   3   3   1

This new triangle is Pascal’s Triangle.

It follows a counting method different from difference triangles.

The sum of the X-th number in the n-th difference  and 
the (X+1)-th number in the n-th difference yields the
(X+1)-th number in the (n-1)-th difference.

It would take a lot of adding if we were to use the difference triangles in the X-gon to

compute (X+1)^10. However, using the Pascal’s Triangle which we have derived from it, the task becomes much simpler. Let’s expand Pascal’s Triangle.

(X+1)^0                                    1
(X+1)^1                                  1   1
(X+1)^2                                1   2   1
(X+1)^3                             1    3   3    1
(X+1)^4                           1    4   6   4    1
(X+1)^5                         1   5   10   10   5   1
(X+1)^6                      1   6   15   20   15   6   1
(X+1)^7                   1   7   21   35   35   21   7    1
(X+1)^8                1   8  28    56   70   56   28   8    1
(X+1)^9              1   9   36   84   126  126  84    36   9    1
(X+1)^10          1   10  45   120   210  252  210   120  45   10    1

The final line of the triangle tells us that

(X+1)^10 = 1X^10 + 10X^9 + 45X^8 + 120X^7 + 210X^6 + 252X^5 + 210X^4 + 120X^3 + 45X^2 + 10X^1 + 1X^0.

Intermediate Algebra/Binomial Theorem

From Wikibooks, the open-content textbooks collection

[edit] Algebra

[edit] The visual way to do the binomial theorem

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