Yimin Math Centre. 2/3 Unit Math Homework for Year 12 (Answers) 3.1 The Binomial Theorem Greatest Coefficient and Greatest Term...

Transcription

1 /3 Unit Math Homework for Year 1 (Answers) Student Name: Grade: Date: Score: Table of contents 3 The Binomial Theorem Part The Binomial Theorem Greatest Coefficient and Greatest Term Identities on the Binomial Coefficients The First Approach - Substitution Second Approach - Differentiation and Integration Third Approach- Equating Coefficients Practical Exam Questions This edition was printed on September 3, 013. Camera ready copy was prepared with the L A TEXe typesetting system. /3 Unit Math Homework for Year 1

2 Year 1 Topic 3 Worked Answers Page 1 of 13 3 The Binomial Theorem Part 3.1 The Binomial Theorem The Binomial Theorem: For all cardinal numbers n, n C r n!, for r 0, 1,..., n r!(n r)! Example Alternatively, n C r n (n 1) (n )... (n r + 1) r The formula for n C r remains unchanged when r is replace by n r: n n! C n r (n r)![n (n r)]! n! (n r)!r! n C r 1. Find the general term in the expansion of (x x 1 ) 0.. Find the term in x 34. General term 0 C r (x ) 0 r ( x 1 ) r 0 C r 0 r x 40 r ( 1) r x r 0 C r 0 r ( 1) r x 40 3r. To obtain the term in x 34, 40 3r 34. r. Hence the term in x 34 0 C (x ) 0 ( x 1 ) x 36 x x x Find the term in x 5. To obtain the term in x 5, 40 3r 5, r 15. Hence the term in x 5 0 C 15 (x ) 0 15 ( x 1 ) x 10 x x x 5.

3 Year 1 Topic 3 Worked Answers Page of 13 Some Particular Values of n C r : For all cardinals n, n C 0 n C n 1, n C 1 n C n 1 n, n C n C n 1 n(n 1). Example 3.1. Find the value of n if: 1. n C 55 We know n C 0 n C n 1,. n C + n C 1 + n C n(n 1) 55 n n n C 1 n C n 1 n, (n 11)(n + 10) 0 since n 0, n 11. n C + n C 1 + n C n(n 1) + n n n + n + 58 n + n 56 0 (n 7)(n + 8) 0 since n 0, n 7. n C n C n 1 n(n 1) Exercise Use the identities: n C r n C n r and n C r 1 + n C r n+1 C r to solve the following equations for n. 1. n C 7 + n C 8 11 C 8 Since n C r 1 + n C r n+1 C r n C 7 + n C 8 n+1 C 8 Now n+1 C 8 11 C 8 n , n C 4 1 C n. Since n C r n C n r 1 C 4 1 C n 4 n 4 or n 4 4 n 8.

4 Year 1 Topic 3 Worked Answers Page 3 of 13 Exercise 3.1. Find the specified terms in each of the following expansions: 1. For the ( + x) 7 : find the term in x 4. Given that the general term is: n C r (a) n r (b) r the term in x 4 is: 7 C 4 () 7 4 x 4 80x 4.. For (x + 1 y)14 : find the term in x 5 y 9, the term in x 5 y 9 is: 14 C 9 (x) 14 9 ( 1 y ) 9 00x y x5 y For (a b 1 ) 0 : find the term in a 3 b 17. the term in a 3 b 17 is: 0 C 17 (a) 0 17 ( b 1 ) a 3 b 17 Exercise Use the particular formulae for n C r for n 0, 1, and, solve each of the following equation for n: 1. n C + 6 C 7 C,. n C + n C 1 n C 0 Given that: n C 0 n C n 1, n C 1 n C n 1 n, 1 (n(n 1) + 1 6(6 1) 1 7(7 1) n n 1 0 n 4. 1 n(n 1) + n + 1, n + n 4 0, n 6.. n C n C n 1 n(n 1) 3. n C 3 + n C 8 n C 1 n C 3 n! 3!(n 3)! n(n 1)(n )(n 3)! 3!(n 3)! 1 n(n 1)(n ). 6 n C 3 + n C 8 n C n(n 1)(n ) + 1 n(n 1) 8n n 3 n n + n + 3n 3n 48n 0 n 3 49n 0, n 7.

5 Year 1 Topic 3 Worked Answers Page 4 of Greatest Coefficient and Greatest Term Definition: In a typical binomial expansion like: (1 + x) x + 4x + 3x x 4, the greatest coefficient is the coefficient of x 3, which is 3. Example 3..1 Write the expansion of (1 + x) 4 in the form t kx k. Find the ratio t k+1 t k. Expanding, (1 + x) 4 So (1 + x) C k (x) k 4 4 C k k x k 4 t k x k, where t k 4 C k k. Hence t k+1 4 C k+1 k+1 t 4 k C k k 4! k!(4 k)! (k + 1)!(3 k)! 4! (4 k) k k k + 1. To find where the coefficients are increasing, we solve t k+1 > t k, That is t k+1 t k From above, > 1 (notice that t k is positive). 8 k k + 1 > 1 8 k > k + 1 k < 1 3. So t k+1 > t k for k 0, 1, and t k+1 < t k Hence t 0 < t 1 < t < t 3 > t 4 and the greatest coeffcient is t 3 4 C for k 3. (k must be an integer). Exercise 3..1 Let ( + 3x) 1 1 t kx k, where t k 1 C k 1 k 3 k is the coefficient of x k. Write down expressions for t k and t k+1 and show that t k+1 t k 36 3k. k+ t 1 k+1 C k+1 1 (k+1) 3 k+1 t 1 k C k 1 k 3 k 1! k!(1 k)! 3 (k + 1)!(11 k)! 1! k!(1 k)(11 k)! 3 (k + 1)k!(11 k)! 36 3k k +.

6 Year 1 Topic 3 Worked Answers Page 5 of 13 Definition: In a typical binomial expansion like: (1 + 3 x) x x x x4, there are two greatest coefficients: x, which is 13 1 and x3 which is Example 3.. Write the expansion of (1 + x) 4 in the form n T k. Find the ratio T k+1 T k, and hence find the greatest term if x 3. 4 From above, (1 + x) 4 T k+1 T k 4 T k, where T k 4 C k k x k. 4 C k+1 k+1 x k+1 4 C k k x k (8 k)x, using the previous working, k + 1 3(8 k) 4(k + 1), substituing x k, after cancelling the s. k + To find where the terms are increasing, we solve T k+1 > T k, that is, T k+1 T k From above, > 1 (again, T k is positive). 1 3k > 1 1 3k > k +, k <. k + So T k+1 > T k for k 0 and 1, T k+1 T k for k, and T k+1 < T k for k 3. Hence T 0 < T 1 < T T 3 > T 4, ( ) ( ) and the greatest terms are T 4 C 13.5, and T 3 4 C Exercise 3.. Let (7 + 3x) 5 5 C kx k. Write down expressions for C k and C k+1, and show that C k+1 C k 75 3k. 7k+7 C k+1 C k 5 C k (k+1) 3 k+1 5 C k 7 5 k 3 k k!(5 k)(4 k)! (k + 1)k!(4 k)! 5 k (k k 7k

7 Year 1 Topic 3 Worked Answers Page 6 of Identities on the Binomial Coefficients Basic Binomial Expansion:. (x + y) n n C k x n k y k The First Approach - Substitution Example Obtain identities by substituting into the basic expansion: 1. x 1 and y 1 We begin with the expansion: (x + y) n substituting x 1 and y 1 gives n n C k x n k y k. n C k, That is. n C 0 + n C 1 + n C n C n n. In the Pascal triangle, this means that the sum of evry row is n.. x 1 and y 1 Substituting x 1 and y 1, gives 0 n C k ( 1) k, That is, n C 0 n C 1 + n C...( 1) n n C n 0. This means that the alternating sum of every row is zero. For odd n, this is trivial: , but for even n: x 1 and y Substituting x 1 and y, gives 3 n n C k k, that is, 1 n C 0 + n C 1 + n C n n C n 3 n.

8 Year 1 Topic 3 Worked Answers Page 7 of Second Approach - Differentiation and Integration Example 3.3. Consider the expansion (1 + x) n n n C k x k. 1. Differentiate the expansion, then substitute x 1 to obtain an identity. Differentiating, n(1 + x) n 1 Substituting x 1, n n 1 k n C k x k 1. k n C k, That is, 0 n C 0 + n C 1 + n C + 3 n C n n C n 0. Taking as an example the row1, 4, 6, 4, 1, Integrate the expansion, then substitute x 1 to obtain an identity. Then give an example of each identity on the Pascal triangle. Integrating, (1 + x) n+1 n + 1 C + n C k x k+1, for som constant C. k + 1 To find the constant C of the integration, substitute x 0, Then 1 n + 1 C + 0 so C 1 (1 + x)n+1, and 1 n n + 1 n + 1 n C k x k+1 k + 1. Substituting x 1, 0 1 n n C k ( 1) k+1, k That is, x + 1 n n C n 1 C 0 + C n 3 + C ( 1)n+1 n C n 0 n + 1 ( 1) n n C n 1 C 0 + C n 3 C ( 1)n n C n 1 n + 1 n + 1 Taking as an example the row 1, 4, 6, 4,

9 Year 1 Topic 3 Worked Answers Page 8 of Third Approach- Equating Coefficients Example Taking ( x + x) 1 n ( ) x + 1 n ( x x + 1 x) n and expanding and equating constants, prove that ( n C 0 ) + ( n C 1 ) + ( n C ) ( n C n ) n C n. Then interpret the identity on the Pascal triangle. The constant term on the RHS is n C n. The constant term on the LHS is the sum of the products: n C 0 n C n + n C 1 n C n 1 + n C n C n n C n n C 0. and because n C n k n C k, by the symmetry of the row, this constant term is ( n C 0 ) + ( n C 1 ) + ( n C ) +,... + ( n C n ). Equating the two conatant terms, ( n C 0 ) + ( n C 1 ) + ( n C ) +...( n C n ) n C n, as required. This means that if the entires of any rwo are squared and added, the sum is the middle enty in the row twice as far down. For example, with the row 1, 3, 3, 1, C 3. and with the row 1, 4, 6, 4, C 4. Exercise By considering the expansion of (1 + x) n (1 + x) n (1 + x) n in two different ways, show that ( n 0) + ( n 1) + ( n ) ( n n) ( n n ). (1 + x) n ( ) n For (1 + x) n the coefficient of x n is n (1 + x) n (1 + x) n [( ) ( ) ( ) ( ) ] [( ) ( ) ( ) ( ] n n n n n n n n + x + x x n + x + x )x n 0 1 n 0 1 n ( )( ) ( )( ) ( )( ) ( )( ) n n n n n n n n the coefficent of x n is: + + +, n 1 n 1 n n 0 ( ) ( ) ( ) n n n ( ) n ( ) n ( ) n since, then the coefficient of x n is k n k 0 1 n ( ) n ( ) n ( ) n ( ) n ( ) n Equating the coefficient of x n gives: n n

10 Year 1 Topic 3 Worked Answers Page 9 of Practical Exam Questions Exercise Assume the identity (1+x) n n n C k x k. By differentiating both sides, and making a suitable substitution for x, show that n. n 1 n k1 k.n C k. (1 + x) n n C k x k,. If n C 5 n C 6, find the value of n. ( ) d dx [(1 + x)n ] d n C k x k dx n(1 + x) n 1 k. n C k x k 1 Let x 1, n. n 1 n C 5 n C 6, k. n C k. n(n 1)(n )(n 3)(n 4) n(n 1)(n )(n 3)(n 4)(n 5) n 5 n Prove that: n C k 1 + n C k n+1 C k k1 RHS n+1 (n + 1)! C k k!(n + 1 k)! LHS n C k 1 + n n! C k (k 1)!(n k + 1)! + n! k!(n k)! [ n! 1 (k 1)!(n k)! n k ] k [ ] n! k + n k + 1 (k 1)!(n k)! k(n k + 1) [ ] n! n + 1 (k 1)!(n k)! k(n k + 1) (n + 1)! k!(n + 1 k)! RHS.

11 Year 1 Topic 3 Worked Answers Page 10 of 13 Exercise Find the term independent of ( 3x 3 x) 8. ( 3x 3 ) 8, T r+1 8 C r (3x 3 ) 8 r ( x x )r 8 C r 3 8 r x 3(8 r) ( ) r (x) r 8 C r 3 8 r x 4 3r r ( ) r 8 C r 3 8 r x 4 4r ( ) r for the term independent of x: 4 4r 0 i.e. r 6.. find the term independent of x in the expansion ( x 1 x ) 9. T 7 8 C 6 3 ( ) (x 1x ) 9, T r+1 9 C r (x) 9 r ( 1x ) r 9 C r 9 r x 9 r ( 1) r (x) r 9 C r 9 r ( 1) r x 9 3r For term indepandent of x: 9 3r 0, i.e. r 3 T 4 9 C 3 6 ( 1) If (1 + x) n n C kx k, find the value of n k1 k C k. (1 + x) n C k x k C 0 x 0 + C 1 x + C x C n x n k1 k c k 1C 1 + C + 3C nc n this implies differentiating both sides of: (1 + x) n C k x k with respect to x n(1 + x) n 1 1C 1 + C x + 3C 3 x nc n x n 1 Let x 1, n() n 1 C 1 + C + 3C nc n x n 1 k C k. k1

12 Year 1 Topic 3 Worked Answers Page 11 of 13 Exercise State the binomial theorem for ( x) n where n is a positive integer. ( ) n 1 + n C 1 x + n C x n C r x r +...x n x 1 + nx + n(n 1) x ! n(n 1)..(n r + 1) x r x n. r!. If k is a positive integer, show that ( n) k approaches 1 as n. ( ) k 1 + k. 1 n n If k is fixed and as n, then 1 n, 1 k(k 1) + ( 1! n ) n k n, 1 3. Show that 1 n! < 1 n 1 for all positive integral n 3. ( n ) n n n k 0 and (1 + 1 n )k n. 1 n(n 1) )...(n r + 1) +...n(n. 1 n! n r! n r n n !.n 1).(n n n r!.n n.n 1 n...n r n n n !.1.(1 1 n ) r!.1.(1 1 n )(1 n )...(1 r 1 ) n n As n, (1 + 1 n )n approches the sum of an infinite series and the sum is greater than. Now n! 1 3..(n 1)n there are n terms and in n 1... there are (n 1) terms. n! > n 1 for all exacpt n when n! n 1 Hence 1 n! < 1 for all integral n 3. n 1 ( 4. Use (3) or otherwise deduce that lim n n n) k where < k < 3. From (3) 1 3! < 1, 1 4! < 1 3, 1 5! < 1 4,... ( ! ) + 1 3! + 1 4! +... < ( ! ) (limiting sum of a geometric series with a 1, and r 1 ) ( ! ) + 1 3! + 1 4! + 1 5! +... < 3 ( 1 + n) 1 n k, where < k < 3. lim n

13 Year 1 Topic 3 Worked Answers Page 1 of 13 Exercise By expanding [x + (1 x)] n, for all real numbers x and all positive integers n, when n is an even number. [x + (1 x)] n n C 0 x n + n C 1 x n 1 (1 x) + n C x n (1 x) n C n (1 x) n Now [x + (1 x)] n [x + 1 x] n 1 n 1 n C 0 x n + n C 1 x n 1 (1 x) + n C x n (1 x) n C n (1 x) n 1. Deduce that n C 0 + n C 1 + n C n C n n. Exercise Substituting x 1 in the above equation, we get: ( n ( ) n 1 ( ) ( ) n ( ) ( ) n n C 0 + ) n C 1 + n C n C n 1 n C 0 n + n C 1 n + n C n +... n C n n 1 n ( n C 0 + n C n C n ) 1 n C 0 + n C 1 + n C +... n C n n. 1. Write the binomial expansion of (1 + x) n. (1 + x) n n C 0 + n C 1 x + n C x + n C 3 x n C n x n. By differentiating this expansion, or otherwise, prove that n r1 r.n C r. r 1 n.3 n 1. Given that : (1 + x) n n C 0 + n C 1 x + n C x + n C 3 x n C n x n Differentiate both sides: n(1 + x) n 1 n C 1 + n C x + 3 n C 3 x n. n C n x n 1 Substitute x : n.3 n 1 n C 1 +. n C n C n. n C n. n 1 r. n C r. r 1 r1 r. n C r. r 1 n.3 n 1 r1

14 Year 1 Topic 3 Worked Answers Page 13 of 13 Exercise Consider the binomial expression: (1 + x) n n C 0 + n C 1 x + n C x + n C 3 x n C n x n. 1. Prove that n C 1 + n C 3 + n C n C n 1 n 1. (1 + x) n n C 0 + n C 1 x + n C x + n C 3 x n C n x n Let x 1, n C 0 n C 1 + n C n C n C n ( 1) n (1 1) 0 n C 1 + n C n C n 1 n C 0 + n C n C n... (A) Let x 1, n n C 0 + n C 1 + n C +... n C n n ( n C 0 + n C n C n ) + ( n C 1 + n C n C n 1 ) Substirute (A), n ( n C 1 + n C n C n 1 ) n 1 n C 1 + n C n C n 1. Prove that n r1 rn C r n n 1. Given that: (1 + x) n n C 0 + n C 1 x + n C x + n C 3 x n C n x n Differnetiate both sides: n(1 + x) n 1 n C 1 + n C x + 3 n C 3 x n. n C n x n 1 Let x 1, n n 1 n C 1 + n C + 3 n C n. n C n r n C r n. n 1. r1 3. Find an expression for n r0 (r + 1)n C r. Given that: (1 + x) n n C 0 + n C 1 x + n C x + n C 3 x n C n x n Multiplying both sides by x: x(1 + x) n n C 0 x + n C 1 x + n C x 3 + n C 3 x n C n x n+1 Differentiating: xn(1 + x) n 1 + (1 + x) n n C 0 + n C 1 x + 3 n C x +... (n + 1) n C n x n Let x 1, n n 1 + n n C 0 + n C n C +...(n + 1) n C n (r + 1) n C r n 1 ( + n). r0