CHAPTER 22 THE BINOMIAL SERIES EXERCISE 93 Page 195 1. Use Pascal's triangle to expand (x – y)7 From Pascal’s triangle on page 195 of the textbook, (a+=++++++ x )6 a 6 6 ax 5 15 ax 42 20 ax 33 15 ax 24 6 ax 5 x 6 Thus, (a+=+++++++ x )7 a 7 7 ax 6 21 ax 52 35 ax 43 35 ax 34 21 ax 25 7 ax 6 x 7 and (xyx− )776 =+ 7()21()35()35()21()7()() xy −+ xy 52 −+ xy 43 −+ xy 34 −+ xy 25 −+−+− xy 6 y 7 i.e. (x−=− y )7 x 7 7 xy 6 + 21 xy 52 − 35 xy 43 + 35 xy 34 − 21 xy 25 + 7 xy 6 − y 7 2. Expand (2a + 3b) 5 using Pascal’s triangle. 5 From page 195 of the textbook, (a+=+++++ x) a55 ax 4 10 ax 32 10 ax 23 5 ax 4 x 5 Thus, replacing a with 2a and x with 3b gives: 5 5 4 32 23 4 5 (2ab+=+++ 3) ( 2 a) 5( 2 ab) ( 3) 10( 2 ab) ( 3) 10( 2 ab) ( 3) ++ 5( 2 ab)( 3) ( 3 b) = 32a5++ 240 ab 4 720 ab32 + 1080 ab23 ++ 810 ab4 243 b 5 345 © 2014, John Bird EXERCISE 94 Page 197 1. Use the binomial theorem to expand (a + 2x)4 4 (43)( ) 2( 432)( )( ) 34 (ax+ 2 ) = a43++42 a( x) a2( 2 x) + ax(2) +( 2 x) 2! 3! = a4++8 a 3 x 24 a 22 x + 32 ax 3 + 16 x 4 2. Use the binomial theorem to expand (2 – x)6 6 (65)( ) 23(654)( )( ) (6)(5)(4)(3) (2 − x) = 26+ 6(2) 5( −+xx) (2)4( −) + (2)3(− x) + (2)24 (− x ) 2! 3! 4! (6)(5)(4)(3)(2) + (2)()()−xx56 +− 5! = 64−+ 192x 240 x2 − 160 x 3456 +−+ 60 x 12 xx 3. Expand (2x – 3y)4 4 4 3 (43)( ) 22(432)( )( ) 3 4 (2xy− 3) =( 2 x) + 42( x) ( − 3 y) + ( 2 x) (− 3 y) + (2x)(− 3 y) +−( 3 y) 2! 3! = 16x4−+ 96 x 3 y 216 x 22 y − 216 xy 3 + 81 y 4 5 2 4. Determine the expansion of 2x + x 5 23 2 54 22(54)( ) 3 (543)( )( ) 2 2 2x+= ( 2 xx) + 52( ) + (2 x) + (2x) x xx2! 3! x 45 (5432)( )( )( ) 22 ++(2x) 4! xx 320 160 32 = 32xxx53+ 160 + 320 +++ xxx35 5. Expand (p + 2q)11 as far as the fifth term. 346 © 2014, John Bird (11)( 10) 23(11)( 10)( 9) (11)(10)(9)(8) (p + 2q)11 = p11++11 p 10 ( 2 q) pq9 (2 ) +pq8 (2 ) + pq7(2 ) 4 2! 3! 4! = p11++22 p 10 q 210 pq9 2 + 1320 pq8 3 + 5280 pq7 4 13 q 6. Determine the sixth term of 3p + 3 n nn( −−1)( n 2) to ( r − 1) terms The 6th term of (ax+ ) is axnr−−( 1) r − 1 (r −1!) 13 5 5 q (13)( 12)( 11)( 10)( 9) 13− 5 q 8 q Hence, the 6th term of 3p + is: (3p) = 1287( 3p) 3 5! 3 3 = 34749 pq85 7. Determine the middle term of (2a – 5b)8 With power 8 there will be 9 terms in the series. The middle term will be the fifth term (2a – 5b)8 (87)( ) 23(876)( )( ) (8)(7)(6)(5) = (2a )87+ 8(2 ab )( −+ 5) (2 ab )6( − 5 ) + (2 ab )5(− 5 ) + (2ab )44 (− 5 ) 2! 3! 4! (8)(7)(6)(5) Hence, the fifth term is: (2ab )44 (− 5 ) = 700000ab44 4! 8. Use the binomial theorem to determine, correct to 4 decimal places: 8 7 (a) (1.003) (b) (0.98) 88(87)( ) ( 876)( )( ) (a) (1.003) =( 1 + 0.003) = 1++ 8(0.003) (0.003)23 +(0.003) + ... 2! 3! = (1+ 0.024 + 0.000252 + 0.000001512 += ...) 1.0242535... = 1.0243, correct to 4 decimal places. 347 © 2014, John Bird (b) 77 (76)( ) 2( 765)( )( ) 3( 7654)( )( )( ) 4 (0.98) =−=+−+( 1 0.02) 1 7( 0.02) ( −+0.02) ( −+0.02) ( −+0.02) ... 2! 3! 4! = 1 – 0.14 + 0.0084 – 0.00028 + 0.0000056 – … = 0.8681, correct to 4 decimal places 9. Evaluate (4.044)6 correct to 2 decimal places. 6 660.044 6 (4.044) =+=+=+( 4 0.044) 4 1 46 ( 1 0.011) 4 6(65)( ) 23( 653)( )( ) ( 7654)( )( )( ) 4 (1+=++ 0.011) 1 6( 0.011) ( 0.011) +(0.011) + (0.011) 2! 3! 4! (7)(6)(5)(4)(3) ++(0.011)5 ... 5! = 1 + 0.066 + 0.001815 + 0.00002662 + 0.000000512435 + 0.000000003382071 = 1.067842136... 6 6 6 + 6 Hence, (4.044) = 4( 1 0.011) = 4 (1.067842136...) = 4373.88, correct to 2 decimal places 348 © 2014, John Bird EXERCISE 95 Page 199 1 1. Expand in ascending powers of x as far as the term in x3, using the binomial theorem. (1− x ) State in each case the limits of x for which the series is valid. 1 −1(−−12)( ) 23( −−− 123)( )( ) =(1 −xx) = 1 +− ( 1)( − ) +( − x) + (− x) +... (1− x) 2! 3! = 1++xx23 + x +... and x < 1 1 2. Expand in ascending powers of x as far as the term in x3, using the binomial theorem. (1+ x ) 2 State in each case the limits of x for which the series is valid. 1 −2(−−23)( ) 23( −−− 234)( )( ) 2 =+(1xx) =−+ 1 2 ( x) + ( x) +... (1+ x) 2! 3! = 1−+ 2xx 323 − 4 x + ... and x < 1 1 3. Expand in ascending powers of x as far as the term in x3, using the binomial theorem. (2+ x )3 State in each case the limits of x for which the series is valid. −3 23 1 −3 x xx(−−34)( ) ( −−− 345)( )( ) x = (2+=x) 2−−33 1 + = 2 1 − 3 + + +... (2+ x )3 2 2 2! 2 3! 2 1 33 5 = 1−+xx23 − x +... 23 22 4 1 33 5 = 1−+xx23 − x +... 8 22 4 x The series is true provided < 1 i.e. x < 2 2 349 © 2014, John Bird 4. Expand 2 + x in ascending powers of x as far as the term in x3, using the binomial theorem. State in each case the limits of x for which the series is valid. 1/2 x x x 2 + x = 21+ = 21+ = 21+ 2 2 2 Using the expansion of (1 + x)n, 1/2 23 x 1xx (1/ 2)(− 1/ 2) (1/ 2)(−− 1/ 2)( 3 / 2) x 21+ = 2 [ ++ +... ] 2 2 2 2! 2 3! 2 xx23 x = 2 1+− + −... 4 32 128 x This is valid when < 1 i.e. x < 2 or –2 < x < 2 2 1 5. Expand in ascending powers of x as far as the term in x3, using the binomial theorem. 13+ x State in each case the limits of x for which the series is valid. 13 135 −− −−− 1 1 − 1 2223 222 = (1+ 3xx) 2 =+− 1 (3 ) + (3 x) + (3 x) ... 13+ x 2 2! 3! 3 27 135 = 1−+xx23 − x as far as the term in x3 2 8 16 1 The series is true provided 3x < 1 i.e. x < 3 6. Expand (2 + 3x)–6 to three terms. For what values of x is the expansion valid? −62 −6 3x 33 xx(−−67)( ) (2+ 3x) = 2−−66 1 + = 2 1 +− ( 6) + to three terms 2 2 2! 2 1 189 1 189 = 1−+ 9xx2 + ... = 1−+ 9xx2 + ... 246 64 4 350 © 2014, John Bird 3x 2 The series is true provided < 1 i.e. x < 2 3 7. When x is very small show that: 1 5 (1− 2x ) 15+ x 19 (a) ≈ 1 + x (b) ≈ 1 + 10x (c) ≈ 1 + x (1−−xx )2 (1 ) 2 (1− 3x ) 4 3 12− x 6 1 1 −−2 x =− −2 ≈+ + (a) 2 (1xx) ( 1) ( 12 x) 1 (11−−xx) ( ) 2 x ≈ 12++x ignoring the x2 term and above 2 5 ≈ 1+ x 2 (12− x) −4 (b) 4 =−−≈−+(12xx)( 13) ( 12 x)( 112 x) (13− x) ≈ 1 + 12x – 2x ignoring the x2 term and above ≈ 1 + 10x 11 15+ x − 5 2 (c) =+(15xx)23( 12 −) ≈+ 1 xx 1 + 3 (12− x) 23 25 ≈ 1++xx ignoring the x2 term and above 32 19 5 2 15+ 4 19 ≈ 1+ x += = 6 23 6 6 8. If x is very small such that x2 and higher powers may be neglected, determine the power series for xx+−483 5 (1+ x ) 3 11 3 11 3 3 xx+−48 −−1 xx 231 =+−(4xxx)23( 8) ( 1 +=+) 5 412 813 −( 1 +x) 5 3 5 (1+ x) 48 3 113xx ≈ ( 4)( 81)+−− ( x) ignoring the x2 term and above 24 38 5 xx3 x 15xx−− 5 72 x 62x ≈ 41+− − ≈ 41+ ≈ 41− 8 24 5 120 120 351 © 2014, John Bird 31x 31 ≈ 41− = 4 – x 60 15 9. Express the following as power series in ascending powers of x as far as the term in x2. State in each case the range of x for which the series is valid. 1− x (1+−xx )3 (1 3 ) 2 (a) (b) 1+ x (1+ x2 ) 11 13 − −− 11 1− x− xx22 2 22 (a) =−(11xx)22( +) ≈ 1 −+( −x) 1 −+ x2 1+ x 2 2! 2 2! as far as the term in x2 xx2 xx332 xxxxx2 22 = 11−− −+ ≈−+ 1 −+ − 28 28 28248 x2 ≈ 1−+x as far as the term in x2 2 The series is valid if x < 1 2 3 21 (1+−xx) ( 13) − (b) =+−+(1xxx)( 13)32( 1 2 ) (1+ x2 ) 21 − 33 2 x2 ≈ (1+xx) 1 −+ 2 ( −3 x) + ..