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# Binomial Theorem

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• 1. The Binomial Theorem By iTutor.com T- 1-855-694-8886 Email- info@iTutor.com
• 2. Binomials  An expression in the form a + b is called a binomial, because it is made of of two unlike terms.  We could use the FOIL method repeatedly to evaluate expressions like (a + b)2, (a + b)3, or (a + b)4. – (a + b)2 = a2 + 2ab + b2 – (a + b)3 = a3 + 3a2b + 3ab2 + b3 – (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4  But to evaluate to higher powers of (a + b)n would be a difficult and tedious process.  For a binomial expansion of (a + b)n, look at the expansions below: – (a + b)2 = a2 + 2ab + b2 – (a + b)3 = a3 + 3a2b + 3ab2 + b3 – (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 • Some simple patterns emerge by looking at these examples: – There are n + 1 terms, the first one is an and the last is bn. – The exponent of a decreases by 1 for each term and the exponents of b increase by 1. – The sum of the exponents in each term is n.
• 3. For bigger exponents  To evaluate (a + b)8, we will find a way to calculate the value of each coefficient. (a + b)8= a8 + __a7b + __a6b2 + __a5b3 + __a4b4 + __a3b5 + __a2b6 + __ab7 + b8 – Pascal’s Triangle will allow us to figure out what the coefficients of each term will be. – The basic premise of Pascal’s Triangle is that every entry (other than a 1) is the sum of the two entries diagonally above it. The Factorial  In any of the examples we had done already, notice that the coefficient of an and bn were each 1. – Also, notice that the coefficient of an-1 and a were each n.  These values can be calculated by using factorials. – n factorial is written as n! and calculated by multiplying the positive whole numbers less than or equal to n.  Formula: For n≥1, n! = n • (n-1) • (n-2)• . . . • 3 • 2 • 1.  Example: 4! = 4  3  2  1 = 24 – Special cases: 0! = 1 and 1! = 1, to avoid division by zero in the next formula.
• 4. The Binomial Coefficient  To find the coefficient of any term of (a + b)n, we can apply factorials, using the formula: n n! n Cr r r! n r ! Blaise Pascal (1623-1662) – where n is the power of the binomial expansion, (a + b)n, and – r is the exponent of b for the specific term we are calculating.  So, for the second term of (a + b)8, we would have n = 8 and r = 1 (because the second term is ___a7b). – This procedure could be repeated for any term we choose, or all of the terms, one after another. – However, there is an easier way to calculate these coefficients. Example : 7 C3 7! 7! 7 (7 3)! • 3! 4! • 3! 4! • 3! (7 • 6 • 5 • 4) • (3 • 2 • 1) (4 • 3 • 2 • 1) • (3 • 2 • 1) 7•6•5• 4 4 • 3 • 2 •1 35
• 5. Recall that a binomial has two terms... (x + y) The Binomial Theorem gives us a quick method to expand binomials raised to powers such as… (x + y)0 (x + y)1 (x + y)2 (x + y)3 Study the following… Row Row Row Row Row Row Row 0 1 This triangle is called Pascal’s 1 Triangle (named after mathematician 1 1 Blaise Pascal). 2 1 2 1 3 1 3 3 1 Notice that row 5 comes from adding up 4 1 4 6 4 1 row 4’s adjacent numbers. (The first row is named row 0). 5 1 5 10 10 5 1 6 1 6 15 20 15 6 1 This pattern will help us find the coefficients when we expand binomials...
• 6. Finding coefficient  What we will notice is that when r=0 and when r=n, then nCr=1, no matter how big n becomes. This is because: n C0 n! n 0 ! 0! n! 1 n! 0! n Cn n! n n ! n! n! 1 0! n!  Note also that when r = 1 and r = (n-1): n C1 n! n 1 ! 1! n n 1! n 1 ! 1! n n Cn 1 n n! n 1 ! n 1! n n 1! 1! n 1 !  So, the coefficients of the first and last terms will always be one. – The second coefficient and next-to-last coefficient will be n. (because the denominators of their formulas are equal) n
• 7. Constructing Pascal’s Triangle  Continue evaluating nCr for n=2 and n=3.  When we include all the possible values of r such that 0≤r≤n, we get the figure below: n=0 0C0 n=1 1C0 1C1 n=2 2C0 n=3 n=4 3C0 4C0 n=5 5C0 n=6 6C0 6C1 3C1 4C1 5C1 2C1 6C2 3C2 4C2 5C2 2C2 4C3 5C3 6C3 3C3 4C4 5C4 6C4 5C5 6C5 6C6
• 8.  Knowing what we know about nCr and its values when r=0, 1, (n-1), and n, we can fill out the outside values of the Triangle: r=n, nCr=1 r=1, nCr=n r=(n-1), nCr=n n=0 1 0C0 n=1 r=0, nCr=1 1 0 1C 1C1 1C1 1 1 n=2 1 1 2 2C 11 C 1 2C01 C2 1 2C2 2 2 n=3 1 0 3C33 33C 111C C2 3 3 31 3C111 C1 3C2 2 3C3 3 n=4 1 0 4CC 44C 44C 111C C3 4 4 4 14 C2 4C111 4 1 4C2 2 4C3 3 4C4 4 n=5 1 0 5C55 55C 55C 55C 111C 54 5 51 2 3 5C111 C1 5C2 2 5C3 3 5C4 4 5C5 5 n=6 1 0 6CC 66C 66C 66C 66C 111C C3 C4 C5 6 6 6 16 C2 6C111 6 1 6C2 2 6C3 3 6C4 4 6C5 5 6C6 6
• 9. Using Pascal’s Triangle  We can also use Pascal’s Triangle to expand binomials, such as (x - 3)4.  The numbers in Pascal’s Triangle can be used to find the coefficients in a binomial expansion.  For example, the coefficients in (x - 3)4 are represented by the row of Pascal’s Triangle for n = 4. x 3 4 4 C0 x 1x 4 4 1 3 0 4 x 4 C1 x 3 3 3 4 6 4 1 3 6 x 1 2 4 C2 x 9 2 4 x 3 1 2 1 4 C3 x 27 1x 4 12x 3 54x 2 108x 81 1x 1 0 3 81 3 4 C4 x 0 3 4
• 10. The Binomial Theorem ( x y)n with nCr x n nx n 1 y  nCr x n r y r  nxy n 1 y n n! (n r )!r !  The general idea of the Binomial Theorem is that: – The term that contains ar in the expansion (a + b)n is n n r n r r ab or n! arbn n r ! r! r – It helps to remember that the sum of the exponents of each term of the expansion is n. (In our formula, note that r + (n - r) = n.) Example: Use the Binomial Theorem to expand (x4 + 2)3. (x 4 2)3 4 3 C0(x ) 3 4 3 1 (x ) 4 2 C1( x ) (2) 3 4 2 C2(x )( 2) 3 3 ( x 4 ) 2 (2) 3 (x 4 )( 2) 2 x12 6 x8 12 x 4 8 1 (2) (2) 3 C3 3 3
• 11. Example: Find the eighth term in the expansion of (x + y)13 .  Think of the first term of the expansion as x13y 0 .  The power of y is 1 less than the number of the term in the expansion. The eighth term is 13C7 x 6 y7. 13 C7 13! 6! • 7! (13 • 12 • 11 • 10 • 9 • 8) • 7! 6! • 7! 13 • 12 • 11 • 10 • 9 • 8 1716 6 • 5 • 4 • 3 • 2 •1 Therefore, the eighth term of (x + y)13 is 1716 x 6 y7.
• 12. Proof of Binomial Theorem  Binomial theorem for any positive integer n, a b n n c0an n c1a n 1b nc2an 2b2 ........ ncnbn Proof The proof is obtained by applying principle of mathematical induction. Step: 1 Let the given statement be f (n) : a b n n c0an n c1an 1b nc2an 2b2 ........ ncnbn Check the result for n = 1 we have f (1) : a b 1 1 c0a1 1c1a1 1b1 a b Thus Result is true for n =1 Step: 2 Let us assume that result is true for n = k f (k ) : a b k k c0ak k c1ak 1b k c2ak 2b2 ........ k ck bk
• 13. Step: 3 We shall prove that f (k + 1) is also true, k 1 f (k 1) : a b k 1 c0ak 1 k 1 c1ak b k 1 c2ak 1b2 ........ k 1ck 1bk Now, a b k 1 (a b)( a b) k k a b c0 a k k c1a k 1b k c2 a k 2b 2 ........ k ck b k From Step 2 k c0 a k 1 1 k c1a k b k c2 a k 1b 2 ........ k ck ab k k k c0 a k c0 a k b k c1a k 1b 2 ........ k ck 1ab k k c1 k c0 a k b k c2 ... by using k 1 c0 1, k cr k cr k 1 k k ck b k 1 c1 a k 1b 2 ..... k ck k ck 1 ab k cr , and k ck 1 k ck b k 1 k 1 ck 1 1
• 14. k 1 c0 a k 1 k 1 c1a k b k 1 c2 a k 1b 2 ........ k 1 ck ab k k 1 ck 1b k  Thus it has been proved that f(k+1) is true when ever f(k) is true,  Therefore, by Principle of mathematical induction f(n) is true for every Positive integer n. 1
• 15. Call us for more Information: 1-855-694-8886 Visit www.iTutor.com The End