Algebra/Binomial Theorem

The notation ' $$n!$$ ' is defined as n factorial.


 * $$n!=n\times(n-1)\times(n-2)\times(n-3)\times\dots\times3\times2\times1$$

0 factorial is equal to 1.
 * $$0!=1$$

Proof of 0 factorial = 1
 * $$n!=n\times(n-1)!$$
 * When n = 1,
 * $$1!=1\times(1-1)!$$
 * $$1=1\times0!$$
 * And thus,
 * $$0!=1$$

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 $$x\ne0$$ 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
 * $$(1+z)^n=\sum_{k=0}^n {n\choose k}z^k$$.

Noting that
 * $$(x+y)^n=\sum_{k=0}^n {n\choose k}x^{n-k}y^k$$

is symmetric in x and y, the identity
 * $${n\choose n-k}={n\choose k}$$

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

A recursive relationship between the $${n\choose k}$$ may be established by considering
 * $$(1+z)^{n+1}=(1+z)(1+z)^n=\sum_{k=0}^{n+1} {n+1\choose k}z^k=(1+z)\sum_{k=0}^n {n\choose k}z^k$$

or
 * $$\sum_{k=0}^{n+1} {n+1\choose k}z^k=\sum_{k=0}^n {n\choose k}z^k+\sum_{k=0}^n {n\choose k}z^{k + 1}=\sum_{k=0}^n {n\choose k}z^k+\sum_{k=1}^{n+1} {n\choose k-1}z^k$$.

Since this must hold for all values of z, the coefficients of $$ z^k $$ on both sides of the equation must be equal
 * $${n+1\choose k}={n\choose k}+{n\choose k-1}$$

for k ranging from 1 through n, and
 * $${n+1\choose n+1}={n\choose n}=\frac{n!}{(n-n)!n!}=\frac{n!}{n!}=1$$
 * $${n+1\choose 0}={n\choose 0}=\frac{n!}{(n-0)!0!}=\frac{n!}{n!}=1$$.

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

Show
 * $${n\choose k}=\frac{n!}{k!(n-k)!}$$

(proof by induction on n).

A useful identity results by setting $$z=1$$
 * $$\sum_{k=0}^n {n\choose k}=2^n$$.

The visual way to do the binomial theorem
(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 =         1x0           =                1 (x+1)1 =       1x1+1x0         =              1   1 (x+1)2 =     1x2+2x1+1x0       =            1   2   1 (x+1)3 =   1x3+3x2+3x1+1x0     =          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 = 1x10 + 10x9 + 45x8 + 120x7 + 210x6 + 252x5 + 210x4 + 120x3 + 45x2 + 10x1 + 1x0.

Practice Problems
 { Problem 1: Find the following without using a pencil and paper:}

{$$99^2$$ { 9801_7 }
 * type="{}"}

{$$998^2$$ { 996004_7 }
 * type="{}"}

{$$101^3$$ { 1030301_7 }
 * type="{}"}

Problem 2: If 3! * 5! * 7! = n!, what is n?