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- 1. If (1+x)2 = (1+x)(1+x) = 1+2x+x2 BINOMIAL EXPANSION How to expand if (2 + 3x)9 = ?????DEFINITION OF BINOMIAL EXPANSIONA sequence have 2 term or more is named as binomial sequence or binomial expression.The exampleare : (a + x), (1 - x), (2x + 3y)Definition 1:“ FACTORIAL” for a number is a multiplication / product of a sequence number.Symbol: !Example : a product of first 7 positive integer are ;7 x 6 x 5 x 4 x 3 x 2 x 1 read as 7 factorial therefore 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1General form is given by the definition below.For any integer positive n, we definite “ FACTORIAL n ” which sign by n! asn! = n(n-1)(n-2)( n-3)………3x2x1 where, 0! = 1Example 1.Find the value of , a) 9! 18! d) b) 6! 15!5! 9! c) 6!Definition 2:Co-eficient of Binomial ; n n n n n!Symbol @ Cr is define as = Cr = r r n r !r!
- 2. We can use the button on our calculator to find these valuesExample 2:Find the value of , 5 20 a) c) 3 17 n 6 5 b) d) 0 4 3EXERCISE 1 EXERCISE 2 ANSWER :Find the value of ; Without using the calculator, find: a) 8! EX. 1: b) 4!7! 5 7! a) a) 40320 c) 2 0! b) 120960 18 c) 5040 d) 6! – 3! b) 8! 3 d) 714 e) e) 56 6! 14! f) 1001 f) g) 1344 10!4! h) 30240 4!8! g) EX. 2: 6! a) 10 h) 8! – ( 2!7!) b) 816
- 3. BINOMIAL THEOREMExpansion in general form (a + b ) n where a, b is any number and n is integer positive is given as n n 0 n n 1 1 n n 2 2 n n r r n 0 n(a + b ) n = a b + a b + a b +……….+ a b +……+ a b 0 1 2 r m or n! a 0b n n! n! n! n! n!(n n)!(a + b ) n = a nb0 + a n 1b1 + a n 2 b 2 +….+ a n r b r +..........+ 0!(n 0)! 1!(n 1)! 2!(n 2)! r!(n r )!Or(a + b)n = an + nC1an – 1b + nC2an – 2b2 + nC3an – 3b3 + ... + nCran – rbr + ... + xnOr we simplify generally as : (a + b ) n = n Cran – rbr …………[ general form]Note: a) Remember !! power of a is desending,power of b is ascending b) First term , T1= an c) Final term T(n)= bn n n r r d) Term of ( r+1) , T( r+1) = a b rExample 3 :By using the binomial theorem,expand all the following; a) ( 3+ x) 5 x 5 b) ( 4-x) 6 d) 2 2 c) ( 1- 3x) 5
- 4. PASCALS TRIANGLE Another way to expand the ( a + b )n , where n is more than 2 For small values the easiest way to determine the value of several consecutive binomial coefficients is with Pascals Triangle:How Pascals Triangle is Constructed1. We start to generate Pascal’s triangle by writing down the number 1. Then we write a new row with the number 1 twice: 1 1 12. We then generate new rows to build a triangle of numbers. Each new row must begin and end with 1: 1 1 1 1 * 1 1 * * 13. The remaining numbers in each row are calculated by adding together the two numbers in the row above which lie above-left and above-right. So, adding the two 1’s in the second row gives 2, and this number goes in the vacant space in the third row: 1 1 1 1 2 1 1 * * 14. The two vacant spaces in the fourth row are each found by adding together the two numbers in the third row which lie above-left and above-right: 1 + 2 = 3, and 2 + 1 = 3. This gives: 1 1 1 1 2 1 1 3 3 1We can continue to build up the triangle in this way to write down as many rows as we wish.Pascal’s trianglen=0 1n=1 1 1n=2 1 2 1n=3 1 3 3 1n=4 1 4 6 4 1n=5 1 5 10 10 5 1Exercise 11. Generate the seventh, eighth, and ninth rows of Pascal’s triangle.
- 5. USING PASCAL’S TRIANGLE TO THE BINOMIAL EXPRESSION. Example: 1. By using the pascal’s triangle, expand ( 2+3x)4 . STEP: - First create the pascal’s triangle….where n=4 n=0 1 n=1 1 1 n=2 1 2 1Take the value in n=3 1 3 3 1this row as a co- n=4 1 4 6 4eficient 1 - Recognize expression of a and b …. So, for ( 2+3x)4 …… 2 is a and 3x is b - Then put it in the normal way as binomial….where the power of a is descrease ( menurun) and power of b is increase ( menaik)….. Result: ( 2+3x)4 = 1 (2)4 (3x)0 + 4 (2)3 (3x)1 + 6 (2)2 (3x)2+ 4 (2)1 (3x)3+ 1 (2)0 (3x)4 See..all the co-efiecient is from the pascal’s triangle..which is 1,4,6,4,1 Example: By using the binomial theorem,expand all the following; a) ( 3+ x) 5 b) ( 4-x) 6 c) ( 1- 3x) 5 d) 5 x 2 2

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