Reget College Maths Departmet Further Pure Proof by Iductio Further Pure Proof by Mathematical Iductio Page
Further Pure Proof by iductio The Edexcel syllabus says that cadidates should be able to: (a) use the method of mathematical iductio to establish a give result (ot restricted to Summatio of series); (b) Recogise situatios where cojecture based o a limited trial followed by iductive proof is a Useful strategy, ad carry this out i simple cases, e.g. to fid the th power of the matrix 0. You eed to apply the followig four steps Basis: show the geeral statemet if true for =. Assumptio: Assume that the geeral statemet is true for =k. Iductio: Show the geeral statemet is true for =k+. Coclusio: The state that the geeral statemet is the true for all positive itegers,. Further Pure Proof by Mathematical Iductio Page 2
Sectio : Geeral Pricipals There are two steps ivolved i provig a result by iductio: Step : Prove true whe =. Step 2: (The iductive step). Assume the result is true for = k ad the prove true for = k +.. Summig series Example: Prove by iductio that 2 r ( )(2 ) for all positive iteger values of. 6 r Solutio: We wish to show that 2 2 2 3 2... 2 ( )(2 ) (*) 6 Step : This is to prove the result true whe = : Left had side of equatio (*) = 2 =. Right had side of equatio (*) = ( )(2 ) =. So equatio (*) is true whe =. Step 2: We assume the result is true whe = k, i.e. we assume that 2 2 2 2 2 3... k ( k )(2k ) We wat to prove the result is true whe = k +, i.e. we wish to show that i.e. 6 k 6 2 2 2 2 2 k 6 2 3... k ( k ) ( k )(2( k ) ) 2 2 2 2 2 k 6 2 3... k ( k ) ( k 2)(2k 3). But 2 2 2 3 2... k 2 ( k ) 2 k ( k )(2k ) ( k ) 2 6 (usig our assumptio). 2 2 2 2 2 ( k) So, 2 3... k ( k ) k(2k ) 6( k ) 6 ( k ) ( k) 2 2 2 2 2 2 2 Therefore, 2 3... k ( k ) 2k k 6k 6 (2k 7k 6) 6 6 Factorisig we get: 2 2 2 2 2 k 6 2 3... k ( k ) ( k 2)(2k 3) as required. Therefore the result is true for = k +. So, by iductio, the result is true for all itegers. Further Pure Proof by Mathematical Iductio Page 3
Examiatio style questio Prove by iductio that r ( ) for all positive iteger values of. 2 r Further Pure Proof by Mathematical Iductio Page 4
Worked examiatio questio (AQA Jauary 2006) a) Prove by iductio that for all itegers. b) Show that 2 2 (3 2) (42 )... ( )2 2 (*) 2 r r ( r)2 2 (2 ) Solutio: Step : This is to prove the result true whe = : Left had side of equatio (*) = ( )2 2 Right had side of equatio (*) = 2 = 2. So equatio (*) is true whe =. Step 2: We assume the result is true whe = k, i.e. we assume that 2 k 2 (3 2) (42 )... ( k)2 k2. We wat to prove the result true whe = k +, i.e. we wat to show that i.e. that 2 k k 2 (3 2) (42 )... ( k)2 ( k )2 2 k k k 2 (32) (42 )... ( k )2 ( k 2)2 ( k )2. k But, So i.e. i.e. So we have 2 k k k k 2 (3 2) (42 )... ( k )2 ( k 2)2 k2 ( k 2)2 (usig the assumptio) 2 k k k 2 (3 2) (42 )... ( k )2 ( k 2)2 2 ( k ( k 2)) 2 k k k 2 (3 2) (42 )... ( k )2 ( k 2)2 2 (2k 2) 2 k k k 2 (32) (42 )... ( k )2 ( k 2)2 2 2( k ) 2 k k k 2 (3 2) (42 )... ( k )2 ( k 2)2 2 ( k ) as required. Therefore the equatio is true whe = k +. So the equatio is true for all iteger values. b) 2 2 r r r ( r )2 ( r )2 ( r )2 r r r Usig the result from part (a), we kow that ad Therefore, 2 r 2 r r 2 ( r)2 2 2 r ( r)2 2. r r 2 ( r )2 2 2 2 2 2 2 2 Further Pure Proof by Mathematical Iductio Page 5
Factorisig gives: 2 r ( r )2 2 (2 2 ) 2 (2 ) as required. r Examiatio questio (OCR Jauary 2005) Prove by iductio that 4 25 36... ( 3) ( )( 5) for all itegers. 3 Examiatio questio (Edexcel) Prove, by iductio, that r 2 2 r ( r ) ( )( )(3 2) Further Pure Proof by Mathematical Iductio Page 6
.2 Sequeces Example: Prove that if u 3u 4 for, ad u 2, the u 43 2. Solutio: Step : We prove the formula true whe = : u 43 2 4 2 2 (which is true). Step 2: Assume the formula is true whe = k, i.e. that We eed to prove the result true whe = k +, i.e. that But, So, u k u k k 3uk 4 343 2 4 k u k u k (usig the assumptio). 3 43 6 4 43 2 as required. Therefore the formula is true whe = k +. Therefore the result is true for all itegers. Examiatio questio (AQA Jue 2005) The sequece u, u2, u 3,... is defied by Prove by iductio that, for all, k 2 u 0, u ( u ). u 2. 2 k 43 2. k 4 3 2 4 3 2. k Further Pure Proof by Mathematical Iductio Page 7
Worked examiatio questio (Edexcel 2005) 6x 0 q (a) Express i the form p +, where p ad q are itegers to be foud. x 3 x 3 6u 0 The sequece of real umbers u, u 2, u 3,... is such that u = 5.2 ad u + =. u 3 (b) Prove by iductio that u > 5, for Z +. Solutio: a) Note that p + Writig q x 3 6x 0 = x 3 = p( x 3) q. x 3 So 6 = p (comparig coefficiets of x) p( x 3) q, we must have 6x + 0 = p(x + 3) + q. x 3 ad 0 = 3p + q i.e. 0 = 8 + q i.e. q = -8. 6x 0 Therefore = 6 - x 3 8 x 3. () (4) b) u + = 6u u 0 3 = 6-8 u 3. Step : Prove the result true whe =, i.e. that u > 5. This is trivially true as u = 5.2. Step 2: Assume true whe = k, i.e. that u k > 5. We the eed to prove the result true whe = k +, i.e. that u k+ > 5. As u k > 5, the u k + 3 > 8 ad so Therefore u + = 6u u 0 3 So the result is true whe = k +. Therefore the result is true for Z +. 8 u 3 <. 8 = 6 - > 6 = 5 (as required) u 3 Further Pure Proof by Mathematical Iductio Page 8
Examiatio questio (NICCEA) Cosider the sequece defied by the relatioship u 5u 2 whose first term is u. (i) Show that the first four terms are, 7, 37, 87, (ii) Use the method of iductio to prove that u 2 3(5 ). Further Pure Proof by Mathematical Iductio Page 9
.3 Divisibility Some questios give you a formula that defies a sequece (i the form u = f() ) ad the ask you to prove that all terms of the sequece are divisible by a particular umber. These questios are usually tackled by simplifyig f(k + ) f(k) OR f(k + ) f(k). Adapted past examiatio questio (adapted MEI): Let f() = 4 2 3. By cosiderig f( + ) f(), or otherwise, prove that of 5 for ay positive iteger. 4 2 3 is a multiple Solutio: Let f() = 4 2 3. Step : Prove f() is a multiple of 5. But f() = 5 2 3 = 35 (which is a multiple of 5). Step 2: We assume f(k) is a multiple of 5. We wat to prove that f(k + ) is a multiple of 5. First we try to simplify: 4( ) f( + ) - f() = 2 4 3 - ( 2 3) = Therefore: f(k + ) f(k) = 4 5 2 k 4k So, f(k + ) = 52 f( k). 45 4 2 2 = 4 4 4 2 (2 ) 5 2 Multiple of 5 So, f(k + ) is a multiple of 5. Assumed to be a multiple of 5 Therefore by iductio, f() must be a multiple of 5 for all positive itegers. Further Pure Proof by Mathematical Iductio Page 0
Worked Examiatio Questio: Edexcel 2002 For Z + prove that 2 3 + 2 + 5 + is divisible by 3, Solutio: Let f() = 2 3 + 2 + 5 + If we here calculate, f( + ) + f() = 2 3(+ ) + 2 + 5 + + + (2 3 + 2 + 5 + ) = 2 3 + 5 + 5 + 2 + 2 3 + 2 + 5 + = 2 3 + 5 + 2 3 + 2 + 5 + 2 + 5 + = 2 3+2 (2 3 + ) + 5 + (5 + ) = 9 2 3+2 + 6 5 + Now we are ready to prove the result. Step : First prove true whe = : 2 3 + 2 + 5 + = 32 + 25 = 57 (which is divisible by 3). Step 2: Assume that the result is true whe = k, i.e. that f(k) is divisible by 3. We showed above that f(k + ) + f(k) = 9 2 3k+2 + 6 5 k+ i.e. that f(k + ) = 9 2 3k+2 + 6 5 k+ f(k) So f(k + ) must be divisible by 3 (as required). Therefore the result is true for Z + Divisible by 3 Divisible by 3 Divisible by 3 (by assumptio) Past examiatio questio (AQA Jauary 2004) The fuctio f is give by 3 3 3 f ( ) ( ) ( 2). a) Simplify, as far as possible, f( + ) f(). b) Prove by iductio that the sum of the cubes of three cosecutive positive itegers is divisible by 9. Further Pure Proof by Mathematical Iductio Page
Further Pure Proof by Mathematical Iductio Page 2
Past examiatio questio (Edexcel 2003) f() = (2 + )7. Prove by iductio that, for all positive itegers, f() is divisible by 4. Further Pure Proof by Mathematical Iductio Page 3
Past examiatio questio (Edexcel) 4 4 f() = 242 3, where is a o-egative iteger. a) Write dow f( + ) f(). b) Prove by iductio that f() is divisible by 5. Further Pure Proof by Mathematical Iductio Page 4
.4 Matrix results Example: a) Fid the matrices b) Predict the value of 2 3 ad 0 0. 0 ad prove your result true by iductio. Solutio a) 2 2 0 0 0 0 3 2 3 0 0 0 0. b) It seems sesible to predict that 0 0. We kow the result is true whe =, 2 ad 3. Suppose ow that it is true whe = k, i.e. that We eed to prove the result true whe = k +, i.e. that But: Therefore: k k k 0 0 0 0 0 k k 0 0 (as required). k k 0 0. k k 0 0. (by assumptio) So the result is true for = k +. Therefore our predictio is true for all positive iteger values of. Further Pure Proof by Mathematical Iductio Page 5
Past examiatio questio (MEI adapted) You are give the matrix (i) (ii) Calculate 2 3 A ad A. 4 A 3. 2 4 Prove by iductio that A 2 whe is a positive iteger. Further Pure Proof by Mathematical Iductio Page 6
Further Pure Proof by Mathematical Iductio Page 7 Past examiatio questio (Edexcel 2002) Prove that 4 9 2 = 3 9 3 whe is a positive iteger.
Past examiatio questio 0 Let A 2. Use iductio to prove that, for all positive itegers, A 0 2 2 2. Further Pure Proof by Mathematical Iductio Page 8
Edexcel Past Examiatio Questios. For Z + prove that (a) 2 3 + 2 + 5 + is divisible by 3, (b) 2 9 3 =. 4 9 3 2. f() = (2 + )7. Prove by iductio that, for all positive itegers, f() is divisible by 4. (9) (7) [P6 Jue 2002 Q 6] (6) 3. (a) Express 6x 0 x 3 i the form p + q x 3, where p ad q are itegers to be foud. [P6 Jue 2003 Q 2] () The sequece of real umbers u, u 2, u 3,... is such that u = 5.2 ad u + = 6u 0. u 3 (b) Prove by iductio that u > 5, for Z +. (4) 4. Prove by iductio that, for Z + r, r2 = 2{ + ( )2 }. r [FP3/P6 Jue 2005 Q ] (5) [*FP3/P6 Jauary 2006 Q 5] 5. Prove by iductio that, for Z + 2, ( 2r ) = 3 (2 )(2 + ). r (5) [FP3 Jue 2007 Q 5] 6. Prove by iductio that, for Z +, r r( r ) =. (5) [FP Ja 2009 Q 4] Further Pure Proof by Mathematical Iductio Page 9
7. A series of positive itegers u, u 2, u 3,... is defied by u = 6 ad u + = 6u 5, for. Prove by iductio that u = 5 6 +, for. (5) [FP Ja 2009 Q 6] 8. Prove by iductio that, for Z +, (a) f() = 5 + 8 + 3 is divisible by 4, 3 (b) 2 2 2 2 =. 2 2 9. A sequece of umbers is defied by (7) (7) [FP Jue2009 Q 8] u = 2, u = 5u 4,. Prove by iductio that, for Z, u = 5 +. 0. f() = 2 + 6. (a) Show that f(k +) = 6f(k) 4(2 k ). (4) [FP Ja 200 Q 3] (3) (b) Hece, or otherwise, prove by iductio that, for Z +, f() is divisible by 8. (4) [FP Jue 200 Q 7]. A sequece of umbers u, u 2, u 3, u 4,..., is defied by u + = 4u + 2, u = 2. Prove by iductio that, for Z +, 2 u = (4 ). 3 (5) [FP Ja 20 Q 9] Further Pure Proof by Mathematical Iductio Page 20
2. Prove by iductio, that for Z +, 3 (a) 6 0 = 3(3 3 0 ) (b) f() = 7 2 + 5 is divisible by 2. 3. A sequece ca be described by the recurrece formula (6) (6) [FP Jue 20Q 9] u = 2u +,, u =. (a) Fid u 2 ad u 3. (b) Prove by iductio that u = 2. 4. Prove by iductio that, for Z +, (2) (5) [FP Ja 202 Q 7] f() = 2 2 + 3 2 is divisible by 5. (6) [FP Jue 202Q 0] 5. (a) Prove by iductio that, for Z +, r (b) A sequece of positive itegers is defied by u =, r( r 3) = ( + )( + 5). 3 (6) u + = u + (3 + ), Z +. Prove by iductio that u = 2 ( ) +, Z +. (5) [FP Ja 203Q 8] Further Pure Proof by Mathematical Iductio Page 2
6. (a) A sequece of umbers is defied by Prove by iductio that, for u = 8 u + = 4u 9,, u = 4 + 3 + (5) (b) Prove by iductio that, for m, 7. (a) Prove by iductio, that for, m 3 4 2m 4m m 2m (5) [FP Jue 203Q 9] (b) Hece, show that r r(2r ) ( )(4 ) 6 (6) 3 r 2 r(2r ) ( a b c) 3 where a, b ad c are itegers to be foud. (4) [FP_R Jue 203Q 8] Further Pure Proof by Mathematical Iductio Page 22