Intermediate Algebra/Binomial Theorem

Algebra

Template:Algebra Page 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

$(1+z{)}^n=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)z^k$ .

Noting that

$(x+y{)}^n=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^{n-k}{y}^k$

is symmetric in x and y, the identity

$\left(\begin{array}{c}n\\ n-k\end{array}\right)=\left(\begin{array}{c}n\\ k\end{array}\right)$

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

A recursive relationship between the $\left(\begin{array}{c}n\\ k\end{array}\right)$ may be established by considering

$(1+z{)}^{n+1}=(1+z)(1+z{)}^n=\sum _{k=0}^{n+1}\left(\begin{array}{c}n+1\\ k\end{array}\right)z^k=(1+z)\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)z^k$

or

$\sum _{k=0}^{n+1}\left(\begin{array}{c}n+1\\ k\end{array}\right)z^k=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)z^k+\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)z^{k+1}=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)z^k+\sum _{k=1}^{n+1}\left(\begin{array}{c}n\\ k-1\end{array}\right)z^k$ .

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

$\left(\begin{array}{c}n+1\\ k\end{array}\right)=\left(\begin{array}{c}n\\ k\end{array}\right)+\left(\begin{array}{c}n\\ k-1\end{array}\right)$

for k ranging from 1 through n, and

$\left(\begin{array}{c}n+1\\ n+1\end{array}\right)=\left(\begin{array}{c}n\\ n\end{array}\right)=1$

$\left(\begin{array}{c}n+1\\ 0\end{array}\right)=\left(\begin{array}{c}n\\ 0\end{array}\right)=1$ .

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

Show

$\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{k!(n-k)!}$

(proof by induction on n).

A useful identity results by setting $z=1$

$\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)=2^n$ .

The visual way to do the binomial theorem

Template:Wikipedia (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.