Algebra/Chapter 10/Binomial Theorem

Algebra Binomial Theorem Quadratic Equation Complex Numbers 10.3: The Binomial Theorem Binomial theorem Factorials The notation ' ' is defined as n factorial. 0 factorial is equal to 1. Proof of 0 factorial = 1 When n = 1, And thus, The Binomial Theorem The binomial thereom gives the coefficients of the polynomial . We may consider without loss of generality the polynomial, of order n, of a single variable z. Assuming set z = y / x . The expansion coefficients of are known as the binomial coefficients, and are denoted . Noting that is symmetric in x and y, the identity may be shown by replacing k by n - k and reversing the order of summation. A recursive relationship between the may be established by considering or . Since this must hold for all values of z, the coefficients of on both sides of the equation must be equal for k ranging from 1 through n, and . Pascal's Triangle is a schematic representation of the above recursion relation ... Show (proof by induction on n). A useful identity results by setting . The visual way to do the binomial theorem 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 = 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. Example Problems Practice Problems {Problem 1: Find the following without using a pencil and paper:} { |type="{}"} { 9801_7 } { |type="{}"} { 996004_7 } { |type="{}"} { 1030301_7 } Problem 2: If 3! * 5! * 7! = n!, what is n?