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© Boardworks Ltd 2005 1 of 39 Contents © Boardworks Ltd 2005 1 of 39 Binomial expansions Geometric sequences Geometric series The sum to infinity of a geometric series Binomial expansions Examination-style questions

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© Boardworks Ltd 2005 2 of 39 Pascal’s Triangle

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© Boardworks Ltd 2005 3 of 39 Binomial expansions An expression containing two terms, for example a + b, is called a binomial expression. When we find powers of binomial expressions an interesting pattern emerges. ( a + b ) 0 = 1 ( a + b ) 1 = 1 a + 1 b ( a + b ) 2 = 1 a 2 + 2 ab + 1 b 2 ( a + b ) 3 = 1 a 3 + 3 a 2 b + 3 ab 2 + 1 b 3 ( a + b ) 4 = 1 a 4 + 4 a 3 b + 6 a 2 b 2 + 4 ab 3 + 1 b 4 ( a + b ) 5 = 1 a 5 + 5 a 4 b + 10 a 3 b 2 + 10 a 2 b 3 + 5 ab 4 + 1 b 5 What patterns do you notice?

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© Boardworks Ltd 2005 4 of 39 Binomial expansions In general, in the expansion of ( a + b ) n : The coefficients are given by the ( n + 1) th row of Pascal’s triangle. The sum of the powers of a and b is n for each term. Altogether, there are n + 1 terms in the expansion. As long as n is relatively small, we can expand a given binomial directly by comparing it to the equivalent expansion of ( a + b ) n. For example: Expand ( x + 1) 5. and replacing a with x and b with 1 gives: Using ( a + b ) 5 = a 5 + 5 a 4 b + 10 a 3 b 2 + 10 a 2 b 3 + 5 ab 4 + b 5 ( x + 1) 5 = x 5 + 5 x 4 + 10 x 3 + 10 x 2 + 5 x + 1

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© Boardworks Ltd 2005 5 of 39 Binomial expansions Expand (2 x – y ) 4. and replacing a with 2 x and b with – y gives: Using ( a + b ) 4 = a 4 + 4 a 3 b + 6 a 2 b 2 + 4 ab 3 + b 4 (2 x – y ) 4 = (2 x ) 4 + 4(2 x ) 3 (– y ) + 6(2 x ) 2 (– y ) 2 + 4(2 x )(– y ) 3 + (– y ) 4 = 16 x 4 – 32 x 3 y + 24 x 2 y 2 – 8 xy 3 + y 4 Notice that when the second term in a binomial is negative the signs of the terms in the expansion will alternate. Suppose we wanted to expand ( a + b ) 20. We could find the 21 st row of Pascal’s triangle, but this would take a very long time.

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© Boardworks Ltd 2005 6 of 39 Finding binomial coefficients When n is large we can find the binomial coefficients using combinations theory. Let’s look more closely at the expansion of ( a + b ) 4 = ( a + b )( a + b )( a + b )( a + b ) aaaa 1 way aaab aaba abaa baaa 4 ways6 ways abbb babb bbab bbba 4 ways bbbb 1 way aabb abab abba bbaa baba baab Ways of getting a 4 Ways of getting a 3 b Ways of getting a 2 b 2 Ways of getting ab 3 Ways of getting b 4

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© Boardworks Ltd 2005 7 of 39 Finding binomial coefficients The situation where no b ’s (or four a ’s) are chosen from any of the four brackets can be written as The situation where one b (or three a ’s) can be chosen from any of the four brackets can be written as: 4 C 0 or. The situation where two b ’s (or two a ’s) can be chosen from any of the four brackets can be written as 4 C 1 or. 4 C 2 or. This is the same as 4 C 4 or. This is the same as 4 C 3 or.

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© Boardworks Ltd 2005 8 of 39 Finding binomial coefficients The fifth row of Pascal’s triangle can be written as: This corresponds to the values 14641 The expansion of ( a + b ) 4 can therefore be written as: Or:( a + b ) 4 = a 4 + 4 a 3 b + 6 a 2 b 2 + 4 ab 3 + b 4

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© Boardworks Ltd 2005 9 of 39 Finding binomial coefficients The number of ways to choose r objects from a group of n objects is written as n C r and is given by n ! is read as ‘ n factorial’ and is the product of all the natural numbers from 1 to n. In general: n ! = n × ( n –1) × ( n – 2) × ( n – 3) … × 2 × 1 n can also be 0 and by definition 0! = 1.

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© Boardworks Ltd 2005 10 of 39 Finding binomial coefficients The value of n! gets large very quickly as the value of n increases. For example: 5! = 5 × 4 × 3 × 2 × 1 =120 12! = 12 × 11 × 10 × … × 2 × 1 =479 001 600 Fortunately, when we use the formula to calculate binomial coefficients, many of the numbers cancel out. For example, for 4 C 2 we have 20! = 20 × 19 × 18 × … × 2 × 1 =2 432 902 008 176 640 000 = 6 2

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© Boardworks Ltd 2005 11 of 39 Finding binomial coefficients Here are some more examples: 56 This value corresponds to the number of ways of choosing 3 a ’s from the 8 brackets in the expansion of ( a + b ) 8. 56 is therefore the coefficient of a 3 b 5 in the expansion of ( a + b ) 8. 36 This value corresponds to the number of ways of choosing 7 a ’s from the 9 brackets in the expansion of ( a + b ) 9. 36 is therefore the coefficient of a 7 b 2 in the expansion of ( a + b ) 9. 4

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© Boardworks Ltd 2005 12 of 39 Finding binomial coefficients The effect of this cancelling gives an alternative form for n C r. In general, the expansion of ( a + b ) n can be written as: A special case is the expansion of (1 + x ) n

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© Boardworks Ltd 2005 13 of 39 Using the binomial theorem Find the coefficient of a 7 b 3 in the expansion of ( a – 2 b ) 10. This method of finding the binomial coefficients is called the binomial theorem. The term in a 7 b 3 is of the form: 4 = 120(–8 a 7 b 3 ) So the coefficient of a 7 b 3 in the expansion of ( a – 2 b ) 10 is –960. = –960 a 7 b 3 3

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© Boardworks Ltd 2005 14 of 39 Using the binomial theorem Use the binomial theorem to write down the first four terms in the expansion of (1 + x ) 7 in ascending powers of x. 3 How could we use this expansion to find an approximate value for 1.1 7 ?

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© Boardworks Ltd 2005 15 of 39 Using the binomial theorem To find an approximate value for 1.1 7 we can let x = 0.1 in the expansion (1 + x ) 7 = 1 + 7 x + 21 x 2 + 35 x 3 + … This gives us We can therefore leave out higher powers of x and still have a reasonable approximation. 1.1 7 ≈ 1 + 0.7 + 0.21 + 0.035 As 0.1 is raised to ever higher powers it becomes much smaller and so less significant. 1.1 7 ≈ 1 +7 × 0.1 +21 × 0.1 2 +35 × 0.1 3 ≈ 1.945

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© Boardworks Ltd 2005 16 of 39 Contents © Boardworks Ltd 2005 16 of 39 Examination-style questions Geometric sequences Geometric series The sum to infinity of a geometric series Binomial expansions Examination-style questions

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© Boardworks Ltd 2005 17 of 39 Examination-style question 2 a)Write down the first four terms in the expansion of (1 + ax ) 13 in ascending powers of x, where a > 0. b)Given that in the expansion of (1 + ax ) 13 the coefficient of x is – b and the coefficient of x 2 is 12 b, find the value of a and b. a) (1 + ax ) 13 = = 1 + 13 ax + 78 a 2 x 2 + 286 a 3 x 3 + … b)13 a = – b 1 78 a 2 = 12 b 2 78 a 2 = 12 × –13 a Substituting into : 12 78 a = –156 a = –2 b = 26

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