Binomial Theorem

1. 2 Binomial Theorem Case Study 2.1 Summation Notation 2.2 Binomial Theorem Chapter Summary

2. Is it possible to find the value of (1.01)n by considering the expansion of (1  0.01)n step by step without using a calculator? You may do this, but if n is large, it will be very time-consuming! Case Study Suppose you are asked to calculate the value of (1.01)n where n is a positive integer, without using a calculator. The student thinks it can be solved by considering the expansion of (1  0.01)n step by step. Do you think this is possible? If the power of the expansion (1  0.01)n is small, we can still find the sum or the values of individual terms by expanding it term by term. Actually, there is an alternative method to tackle such a problem, which is especially effective when the power is large. This method will be discussed in this chapter.

3. Note: Besides i, we can also use other letters such as j, k, m, etc, in the summation notation such as or . general term of the sequence i changes from 1 to n 2.1 Summation Notation In mathematics, a simple way to denote the sum of the terms of a sequence is by using summation notation. It involves the symbol ‘Σ’ which is a Greek letter read as ‘sigma’. For example, consider a sequence with general term T(n). The series T(1) + T(2) + ... + T(n) can be represented by the notation: The notation is read as ‘the summation of T(i) as i goes from 1 to n’.

4. Evaluate the following sums. (a) (b) 2.1 Summation Notation Example 2.1T Solution: (a) (b)

5. Use summation notation to express the series . Let T(n) be the general term of the sequence 1, . 2.1 Summation Notation Example 2.2T Solution: Note that … ∴ ∴

6. coefficients 2.2 Binomial Theorem A. Pascal’s Triangle When we expand the expression (xy)n for a positive integer n, it is too troublesome to use long multiplication, especially when n is very large. Consider (xy)3  (xy)2(xy)  (x2 2xyy2)(xy) ∴ (xy)3x3 3x2y 3xy2y3 Using long multiplication again, (xy)4  (xy)3(xy)  (x3 3x2y 3xy2y3)(xy) ∴ (xy)4x4 4x3y 6x2y2 4xy3y4

7. 1 3 3 1 1 4 6 4 1 2.2 Binomial Theorem A. Pascal’s Triangle The expansion (xy)n follows some special pattern. Binomial Expansion (xy)0 (xy)1 (xy)2 (xy)3 (xy)4 (xy)5 Coefficientsoftheexpansion 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 The above triangle is called Pascal’s Triangle. Besides 1, every number equals the sum of the two numbers to the left and right of it in the preceding row. For example:

8. 2.2 Binomial Theorem A. Pascal’s Triangle In general, besides the above relationship between the coefficients, for n 2, the properties of the expansion of (xy)n are stated as follows: 1. There are (n 1) terms. 2. When the powers of x in the expansion decrease by 1, the powers of y will increase by 1. 3. The sum of the powers of x and y in each term must be equal to n. 4. The coefficients of the first term and the last term are both ‘1’. 5. The coefficients show a symmetrical pattern.

9. 2.2 Binomial Theorem A. Pascal’s Triangle Example 2.3T Using the properties of Pascal’s Triangle, expand the following expressions. (a) (2 ax)4 (b) (x2 1)5 Solution: (a) (2 ax)4 (b) (x2 1)5

10. 2.2 Binomial Theorem B. Binomial Theorem Pascal’s Triangle is very useful for expanding (xy)n when n is small. However, in the case when n is large, such as 30 or more, the work will be tedious. Recall the definition of the notation Cn: r Rewrite Pascal’s Triangle using the notation:

11. Binomial Theorem If n is a positive integer, then (xy)n  Using mathematical induction, we can prove the binomial theorem, that is, (xy)n  for all positive integers n. 2.2 Binomial Theorem B. Binomial Theorem We can see that all the coefficients in Pascal’s Triangle can be rewritten in the form Cn. r The expansion of (xy)n can be obtained by the following binomial theorem:

12. 2.2 Binomial Theorem B. Binomial Theorem Example 2.4T Expand (5x 3)5 in ascending powers of x as far as the 4th term. Solution:

13. (a) Find the 5th term in the expansion of (53x)6 in ascending powers of x. (b) Find the term containing x2 in the expansion of (27x)5. (c) Find the constant term in the expansion of . 2.2 Binomial Theorem B. Binomial Theorem Example 2.5T Solution: (a) The 5th term (c) General term For the constant term, the power of x is 0. i.e., 484r  0 (b) The term in x2 r  12 ∴ Constant term

14. 2.2 Binomial Theorem B. Binomial Theorem Example 2.6T It is given that (2  3x)n 64  18a5x terms involving higher powers of x. Find the values of n and a, where n is a positive integer. Solution: (2  3x)n ∴ 2n 64 n  6

15. 2.2 Binomial Theorem B. Binomial Theorem Example 2.7T (a) Expand(1x)5(5mx)3inascendingpowersof xasfaras the termin x2. (b) If thecoefficientof x2 in theexpansion in(a)is560, find thevaluesof m. Solution: (a) (1x)5(5mx)3 (b) ∵ Coefficient of x2 560 ∴

16. Chapter Summary 2.1 Summation Notation The sum of n terms of a series with the general term T(n) can be denoted by where n is a positive integer.

17. Binomial Theorem If n is a positive integer, then (xy)n Chapter Summary 2.2 Binomial Theorem Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1               