Binomial Theorem
- 1. The Binomial Theorem By iTutor.com T- 1-855-694-8886 Email- [email protected]
- 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
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