Proof by induction

1 24 +31−2=45 is divisible by 15 B1 Checks base case. Assume that 24k +31k −2 is divisible by 15 for some positive integer k. B1 States inductive hypothesis. Then 24k+4 +31k+1−2=(15+1)24k +(30+1)31k −2 M1 A1 Separates 24k +31k −2 or considers difference. is divisible by 15 because 15×24k +30×31k is divisible by 15. A1 Hence, by induction, true for every positive integer n. A1 6 Question Answer Marks Guidance

1 24 +31−2=45 is divisible by 15 B1 Checks base case. Assume that 24k +31k −2 is divisible by 15 for some positive integer k. B1 States inductive hypothesis. Then 24k+4 +31k+1−2=(15+1)24k +(30+1)31k −2 M1 A1 Separates 24k +31k −2 or considers difference. is divisible by 15 because 15×24k +30×31k is divisible by 15. A1 Hence, by induction, true for every positive integer n. A1 6 Question Answer Marks Guidance

3(a) 5×14 +12 = 1 (2)2 ( 2+1 ) [ =6 ] so H is true. B1 Checks base case. 2 1 Assume that  k   ( 5r4 +r2 )   = 1 2 k2(k+1)2 ( 2k+1 ) B1 S th t e a t a e l s g i e n b d r u ai c c t i f v o e r m hy . p I o f t s h a e y s s i s f o [f r o A r L so L m k e , k th ] e in n c B lu 0 d . ing r=1  k+1  ( 5r4 +r2 ) = 1 k2(k+1)2 ( 2k+1 )+5(k+1)4 +(k+1)2 M1 Considers sum to k +1.   2 r=1 ( ) 1(k+1)2 2k3 +k2 +10(k+1)2 +2 M1 Take out the factor of (cid:4666)𝑘+1(cid:4667)(cid:2870) OR expands the 2 summation expression and the target expression for 𝑘+1 and collects like terms for both. 1(k+1)2(2k3+11k2 +20k+12)= 1(k+1)2(k+2)2(2k+3) A1 Factorises or having expanded, checks explicitly. 2 2 At least one intermediate step seen following the award of M1 before reaching the answer. So H is true. By induction, H is true for all positive integers n. A1 States conclusion. Implication must be clearly k+1 n expressed. 6 3(b) n M1 Uses correct formula for r2. 5r4 +1n(n+1)(2n+1)= 1n2(n+1)2(2n+1) 6 2 r=1 [ ] n M1 Makes r4 the subject and takes out all linear 5 r4 = 1n(n+1)(2n+1)(3n(n+1)−1) 6 factors and the remaining term is of correct form. r=1 n A1 CAO r4 = 1 n(n+1)(2n+1)(3n2 +3n−1) 30 r=1 3 © UCLES 2021 Page 9 of 16

3(a) a =23 leading to lna =3ln2 so true when n=2. B1 Checks base case. Must be n =2 and exact. 2 2 Assume that lna ≥3k−1ln2. B1 States inductive hypothesis. k  1  M1 Applies result given in part (a). lna k+1 =3lna k + a  >3lna k   k ≥ 3k ln 2 M1 A1 Applies inductive hypothesis and writes in required form. So true when n=k+1. By induction, true for all integers n ≥ 2. A1 States conclusion, all previous marks must be gained. 6 3(b)  1  M1 Applies results given in parts (a) and (b). 3lna +  −lna >2lna n a n n   n >2×3n−1ln2=3n−1ln4 A1 AG 2 Question Answer Marks Guidance

2 dy M1 A1 Differentiates once using the product rule. =axeax +eax =(ax+1)eax so true when n=1. dx dky ( ) B1 States inductive hypothesis. Assume that = akx+kak−1 eax. dxk dk+1y ( ) ( ) ( ) M1 A1 Differentiates kth derivative. =a akx+kak−1 eax +eax ak = ak+1+(k+1)ak eax dxk+1 So true when n=k +1. By induction, true for all positive integers n. A1 States conclusion. 6 Question Answer Marks Guidance

3(a) a =23 leading to lna =3ln2 so true when n=2. B1 Checks base case. Must be n =2 and exact. 2 2 Assume that lna ≥3k−1ln2. B1 States inductive hypothesis. k  1  M1 Applies result given in part (a). lna k+1 =3lna k + a  >3lna k   k ≥ 3k ln 2 M1 A1 Applies inductive hypothesis and writes in required form. So true when n=k+1. By induction, true for all integers n ≥ 2. A1 States conclusion, all previous marks must be gained. 6 3(b)  1  M1 Applies results given in parts (a) and (b). 3lna +  −lna >2lna n a n n   n >2×3n−1ln2=3n−1ln4 A1 AG 2 Question Answer Marks Guidance

3(a) u 4 (given) B1 States base case. 1 Assume that u 4 for some positive integer k. B1 States inductive hypothesis. k u 2 u 12 u 2 7u 12 M1 Considers u 4, puts over common Then u 4 k k 4 k k k1 k1 2u 2u denominator. k k u 3u 4 A1 u 4  k k 0 k1 2u k Hence, by induction, u 4 for all positive integers n. A1 n 5 3(b) u 2 u 12 u 4(u 3) M1 A1 Considers u u . u u  n n  n n n1 n n1 n 2u 2u n n So u 4u u 0 A1 Uses u 4. n n1 n n 3 © UCLES 2022 Page 8 of 15

3(a) u 4 (given) B1 States base case. 1 Assume that u 4 for some positive integer k. B1 States inductive hypothesis. k u 2 u 12 u 2 7u 12 M1 Considers u 4, puts over common Then u 4 k k 4 k k k1 k1 2u 2u denominator. k k u 3u 4 A1 u 4  k k 0 k1 2u k Hence, by induction, u 4 for all positive integers n. A1 n 5 3(b) u 2 u 12 u 4(u 3) M1 A1 Considers u u . u u  n n  n n n1 n n1 n 2u 2u n n So u 4u u 0 A1 Uses u 4. n n1 n n 3 © UCLES 2022 Page 8 of 15

2 72 +97−50=96 is divisible by 48 B1 Checks base case. Assume that 72k +97k −50 is divisible by 48 for some positive integer k. B1 States inductive hypothesis. Then 72k+2 +97k+1−50=(48+1)72k +(96+1)97k −50 M1 A1 Separates 72k +97k −50 or considers difference. is divisible by 48 because 4872k +9697k is divisible by 48. A1 Hence, by induction, true for every positive integer n. A1 6 Question Answer Marks Guidance

4 d B1 Checks base case using product rule. (xf(x))=xf '(x)+ f(x)=xf '(x)+(2(1)−1)f(x) dx d2k−1 B1 States inductive hypothesis. Assume true for n=k, so (xf(x))=xf '(x)+(2k−1)f(x) dx2k−1 d2k M1 A1 Differentiates once. Then (xf(x))=xf(x)+2kf '(x) dx2k d2k+1 M1 A1 Differentiates again. (xf(x))=xf '(x)+ f(x)+2kf(x) dx2k+1 =xf '(x)+(2k+1)f(x) So, it is also true for n=k+1. Hence, by induction, true for all positive integers. A1 States conclusion. 7 © UCLES 2022 Page 9 of 15

2 72 +97−50=96 is divisible by 48 B1 Checks base case. Assume that 72k +97k −50 is divisible by 48 for some positive integer k. B1 States inductive hypothesis. Then 72k+2 +97k+1−50=(48+1)72k +(96+1)97k −50 M1 A1 Separates 72k +97k −50 or considers difference. is divisible by 48 because 4872k +9697k is divisible by 48. A1 Hence, by induction, true for every positive integer n. A1 6 Question Answer Marks Guidance

1(a) 6 0 23 0 B1 States base case. 2A   so true when n1. 2 2 31 2 23k 0 B1 States inductive hypothesis. Assume that it is true for nk, so 2Ak  .  3k 1 2  23k 03 0  23k1 0 M1A1 Multiplies 2Ak with A. Then 2Ak1      3k 1 2  1 1   3k132 2  So, it is also true for nk1. Hence, by induction, true for all positive integers. A1 States conclusion. 5 1(b)  3n 0 B1 detAn det 3n 1  3n 1  1 2   2 0  Or a multiple of An 213n  seen. 13n 23n   1 0  B1  2 0  An 3n  OE An 213n   1  13n 3n 13n 23n  2  2 © UCLES 2023 Page 6 of 16

1(a) 6 0 23 0 B1 States base case. 2A   so true when n1. 2 2 31 2 23k 0 B1 States inductive hypothesis. Assume that it is true for nk, so 2Ak  .  3k 1 2  23k 03 0  23k1 0 M1A1 Multiplies 2Ak with A. Then 2Ak1      3k 1 2  1 1   3k132 2  So, it is also true for nk1. Hence, by induction, true for all positive integers. A1 States conclusion. 5 1(b)  3n 0 B1 detAn det 3n 1  3n 1  1 2   2 0  Or a multiple of An 213n  seen. 13n 23n   1 0  B1  2 0  An 3n  OE An 213n   1  13n 3n 13n 23n  2  2 © UCLES 2023 Page 6 of 16

1 533233124 is divisible by 31. B1 Checks base case. Assume that 53k 32k 33 is divisible by 31 for some positive integer B1 States inductive hypothesis. k. 53k332k133(1241)53k (311)32k 33 M1 A1 Separates 53k 32k 33 or considers difference. is divisible by 31 because 12453k 3132k is divisible by 31. A1 Hence, by induction, true for every positive integer n. A1 6 Question Answer Marks Guidance

2 1−2x+x2 (1−x)2 B1 Checks base case. 1= = so H is true. (1−x)2 (1−x)2 1 k 1−(k+1)xk +kxk+1 B1 States inductive hypothesis. Assume that rxr−1 = . (1−x)2 r=1 k+1 1−(k+1)xk +kxk+1 M1 Considers sum to k+1. rxr−1= +(k+1)xk (1−x)2 r=1 1−(k+1)xk +kxk+1+(k+1)xk( 1−2x+x2) M1 Puts over a common denominator. (1−x)2 1+kxk+1+(k+1)xk( −2x+x2) 1−(k+2)xk+1+(k+1)xk+2 A1 = (1−x)2 (1−x)2 So H is true. By induction, H is true for all positive integers n. A1 States conclusion. k+1 n 6 © UCLES 2023 Page 7 of 15

2 d ( x2ex) =x2ex +2xex = ( x2 +2x ) ex so true when n=1. M1 A1 Differentiates once using the product rule. dx Assume that dk ( x2ex) = ( x2 +2kx+k(k−1) ) ex B1 States inductive hypothesis. dxk [for some value of k]. dk+1 ( x2ex) = ( x2 +2kx+k(k−1) ) ex +ex(2x+2k) M1 Differentiates kth derivative. dxk+1 ( x2+2(k+1)x+k(k+1) ) ex A1 So true when n=k+1. By induction, true for all positive integers n. A1 States conclusion. 6 Question Answer Marks Guidance

2 1−2x+x2 (1−x)2 B1 Checks base case. 1= = so H is true. (1−x)2 (1−x)2 1 k 1−(k+1)xk +kxk+1 B1 States inductive hypothesis. Assume that rxr−1 = . (1−x)2 r=1 k+1 1−(k+1)xk +kxk+1 M1 Considers sum to k+1. rxr−1= +(k+1)xk (1−x)2 r=1 1−(k+1)xk +kxk+1+(k+1)xk( 1−2x+x2) M1 Puts over a common denominator. (1−x)2 1+kxk+1+(k+1)xk( −2x+x2) 1−(k+2)xk+1+(k+1)xk+2 A1 = (1−x)2 (1−x)2 So H is true. By induction, H is true for all positive integers n. A1 States conclusion. k+1 n 6 © UCLES 2023 Page 7 of 15

2 64 3821332 is divisible by 74. B1 Checks base case. Assume that 64k 38k 2 is divisible by 74 for some positive integer k. B1 States inductive hypothesis. Then 64k4 38k12(12951)64k (371)38k 2 M1 A1 Separates 64k 38k 2 or considers difference. Is divisible by 74 because 129564k 3738k is divisible by 74. A1 Hence, by induction, 64k 38k 2 is divisible by 74, is true for every positive integer A1 n. 6 Question Answer Marks Guidance

2 64 3821332 is divisible by 74. B1 Checks base case. Assume that 64k 38k 2 is divisible by 74 for some positive integer k. B1 States inductive hypothesis. Then 64k4 38k12(12951)64k (371)38k 2 M1 A1 Separates 64k 38k 2 or considers difference. Is divisible by 74 because 129564k 3738k is divisible by 74. A1 Hence, by induction, 64k 38k 2 is divisible by 74, is true for every positive integer A1 n. 6 Question Answer Marks Guidance

2 d ( tan−1x ) 1 B1 Differentiates once. = so true when n=1. dx 1+x2 Assume that d d x k k ( tan−1x ) =P k (x)( 1+x2)−k , where degP k (x)=k−1. B1 S d t e a g te P s k i ( n x d ) u = ct k iv − e 1 h . y pothesis. Must have dk+1( tan−1x ) M1 A1 Differentiates kth derivative using the product =P '(x)( 1+x2)−k −2kxP (x)( 1+x2)−k−1 rule. dxk+1 k k = ( P '(x)( 1+x2) −2kxP (x) )( 1+x2)−k−1 so degP (x)=k A1 Writes in the form P (x)( 1+x2)−k−1 . k k k+1 k+1 So true when n=k+1. By induction, true for all positive integers n. A1 Attempts to show degree of P (x) is at most k k+1 (condone not showing coefficient of xk is non- zero) and states conclusion. 6 © Cambridge University Press & Assessment 2024 Page 7 of 15

1 u =4=31+1 B1 Shows base case 1 Assume that it is true for n=k, so u =3k +1. B1 States inductive hypothesis. k Then u =3(3k +1)−2=3k+1+1 M1 A1 Substitutes into recursion formula. k+1 [So, it is also true for n=k+1]. Hence, by induction, u =3n +1for all positive integers. A1 States conclusion. n 5 Question Answer Marks Guidance

2 d ( tan−1x ) B1 Differentiates once. 1 = so true when n=1. dx 1+x2 Assume that d d x k k ( tan−1x ) =P k (x)( 1+x2)−k , where degP k (x)=k−1. B1 S d t e a g te P s k i ( n x d ) u = ct k iv − e 1 h . y pothesis. Must have dk+1( tan−1x ) M1 A1 Differentiates kth derivative using the product =P '(x)( 1+x2)−k −2kxP (x)( 1+x2)−k−1 rule. dxk+1 k k = ( P '(x)( 1+x2) −2kxP (x) )( 1+x2)−k−1 so degP (x)=k A1 Writes in the form P (x)( 1+x2)−k−1 . k k k+1 k+1 So true when n=k+1. By induction, true for all positive integers n. A1 Attempts to show degree of P (x) is at most k k+1 (condone not showing coefficient of xk is non- zero) and states conclusion. 6 © Cambridge University Press & Assessment 2024 Page 7 of 15

3(a) u =5=61−1 B1 Shows base case. 1 Assume that it is true for n=k, so u =6k −1. B1 States inductive hypothesis. k Then u =6(6k −1)+5=6k+1−1. M1 A1 Substitutes into recursion formula. k+1 So, it is also true for n=k+1. Hence, by induction, u =6n −1is true for all positive A1 States conclusion. n integers. 5 3(b) u 62n −1 ( 6n −1 )( 6n +1 ) M1 A1 Uses (a) and difference of two squares. 2n = = =6n +1 u 6n −1 6n −1 n 2 Question Answer Marks Guidance

3(a) u =5=61−1 B1 Shows base case. 1 Assume that it is true for n=k, so u =6k −1. B1 States inductive hypothesis. k Then u =6(6k −1)+5=6k+1−1. M1 A1 Substitutes into recursion formula. k+1 So, it is also true for n=k+1. Hence, by induction, u =6n −1is true for all positive A1 States conclusion. n integers. 5 3(b) u 62n −1 ( 6n −1 )( 6n +1 ) M1 A1 Uses (a) and difference of two squares. 2n = = =6n +1 u 6n −1 6n −1 n 2 Question Answer Marks Guidance

2 2025+47−2=2070=4546is divisible by 46. B1 Checks base case. Assume that 2025k +47k −2 is divisible by 46 for some positive integer k. B1 States inductive hypothesis. Then 2025k+1+47k+1−2=(2024+1)2025k +(46+1)47k −2 M1 A1 Separates 2025k +47k −2 or considers difference. is divisible by 46 because A1 Shows that it is divisible by 46. 20242025k +4647k = is divisible by 46 Hence, [by induction], true for every positive integer n. A1 Must reference the statement to be proven. 6 Question Answer Marks Guidance

2 d M1 d (xlnx)=1+lnx Finds (xlnx). dx dx d2 A1 Checks base case. (xlnx)=x−1 =(−1)2(2−2)!x1−2 dx2 Allow d2 (xlnx)=x−1 =(−1)2(0)!x1−2 dx2 dk B1 States inductive hypothesis. Assume that (xlnx)=(−1)k(k−2)!x1−k dxk dk+1 M1 Differentiates kth derivative. (xlnx)=(−1)k(k−2)!(1−k)x1−k−1 dxk+1 =(−1)k+1((k+1)−2)!x1−(k+1) A1 dn A1 States conclusion. So, (xlnx)=(−1)n(n−2)!x1−n is also true for n=k+1. dxn [Hence, by induction], true for every integer n 2. 6 © Cambridge University Press & Assessment 2025 Page 10 of 18

3 d B1 Checks base case using product rule. (xcosx)=cosx−xsinx=(−1)1(xsinx−(2(1)−1)cosx) dx d2k−1 B1 States inductive hypothesis. Assume true for n=k, so (xcosx)=(−1)k(xsinx−(2k−1)cosx) dx2k−1 d2k M1 A1 Differentiates once. Then (xcosx)=(−1)k(xcosx+2ksinx) dx2k d2k+1 M1 A1 Differentiates again. (xcosx)=(−1)k(−xsinx+cosx+2kcosx) dx2k+1 =(−1)k+1( xsinx−(2k+1)cosx ) So, it is also true for n=k+1. Hence, [by induction], A1 States conclusion. d2n−1 (xcosx)=(−1)n(xsinx−(2n−1)cosx) is true for all positive integers. dx2n−1 7 Question Answer Marks Guidance

3(a) u 5 (given) B1 States base case. 1 Assume that u 5 for some positive integer k. B1 States inductive hypothesis. k 6u +5 5u +10−6u −5 M1 Considers 5−u , puts over common Then 5−u =5− k = k k k+1 k+1 u +2 u +2 denominator. k k 5−u A1 = k 0. u +2 k Hence, by induction, u 5 for all positive integers n. A1 n 5 3(b) 6u +5 6u +5−u 2 −2u −u 2 +4u +5 M1 A1 Considers u n+1 −u n or uses u −u = n −u = n n n = n n n+1 n u n +2 n u n +2 u n +2 5−u = 5−u n . n+1 u +2 n (5−u )(1+u ) A1 Uses u 5. = n n . So u 5u −u 0 n n n+1 n u +2 n 3 Question Answer Marks Guidance

3 d B1 Checks base case using product rule. (xcosx)=cosx−xsinx=(−1)1(xsinx−(2(1)−1)cosx) dx d2k−1 B1 States inductive hypothesis. Assume true for n=k, so (xcosx)=(−1)k(xsinx−(2k−1)cosx) dx2k−1 d2k M1 A1 Differentiates once. Then (xcosx)=(−1)k(xcosx+2ksinx) dx2k d2k+1 M1 A1 Differentiates again. (xcosx)=(−1)k(−xsinx+cosx+2kcosx) dx2k+1 =(−1)k+1( xsinx−(2k+1)cosx ) So, it is also true for n=k+1. Hence, [by induction], A1 States conclusion. d2n−1 (xcosx)=(−1)n(xsinx−(2n−1)cosx) is true for all positive integers. dx2n−1 7 Question Answer Marks Guidance

1 LHS= (6)1 = 6 RHS=1+1 = 6 B1 Checks base case. 5 5 5 5 Assume that (6)k  1+1k for some positive integer k. B1 States inductive hypothesis. 5 5 Then (6)k+1  (6)( 1+1k ) M1 Moving to the next term and using the inductive 5 5 5 hypothesis. = 6+ 6 k  6+1k =1+ 1(k+1) A1 Convincing algebra to reach the required form. 5 25 5 5 5 Hence, by induction, (6)n  1+1nis true for every positive integer n. A1 Must follow M1 A1. 5 5 5 Question Answer Marks Guidance

2 EITHER © Cambridge University Press & Assessment 2026 Page 10 of 23 3 6 + 9 2 + 8 9 = 9 1 0 (B1 Checks base case. is divisible by 91. Assume that 36k + 92k + 89 is divisible by 91 for some positive integer k. B1 States inductive hypothesis, making it clear it is an assumption. B0 if assume true for ALL values of k. Then 3 6 (k + 1 ) + 9 2 k + 1 + 8 9 = ( 7 2 8 + 1 ) 3 6 k + ( 9 1 + 1 ) 9 2 k + 8 9 M1 = 7 2 8  3 6 k + 9 1  9 2 k + ( 3 6 k + 9 2 k + 8 9 ) A1 Separates 36k + 92k +89 A1 Connects the divisibility with the is divisible by 91 because both 72836k +9192k and are divisible by 91. hypothesis. Hence, [by induction,] 3 6 n + 9 2 n + 8 9 is divisible by 91for every positive integer n. A1) Must have previous three marks and state the result in full. OR Alternative method for 2 3 6 + 9 2 + 8 9 = 9 1 0 is divisible by 91. (B1 Checks base case. Assume that ( 3 6 k + 9 2 k + 8 9 ) is divisible by 91 for some positive integer k. B1 States inductive hypothesis, making it clear it is an assumption. B0 if assume true for ALL values of k. Then 36(k+1) +92k+1+89−(36k +92k +89) M1 Considers difference between consecutive terms. = = ( 9 7 1 2 ( 8 8 )  3 3 6 k 6 k + + ( 9 9 1 2 ) k 9 ) 2 k 36k +92k +89 A1 A1 Connects the divisibility with the hypothesis. Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2 Hence, [by induction,] © Cambridge University Press & Assessment 2026 Page 11 of 23 3 6 n + 9 2 n + 8 9 is divisible by 91 for every positive integer n. A1) Must have previous three marks and state the result in full. 6 Question Answer Marks Guidance

2 EITHER © Cambridge University Press & Assessment 2026 Page 10 of 23 3 6 + 9 2 + 8 9 = 9 1 0 (B1 Checks base case. is divisible by 91. Assume that 36k + 92k + 89 is divisible by 91 for some positive integer k. B1 States inductive hypothesis, making it clear it is an assumption. B0 if assume true for ALL values of k. Then 3 6 (k + 1 ) + 9 2 k + 1 + 8 9 = ( 7 2 8 + 1 ) 3 6 k + ( 9 1 + 1 ) 9 2 k + 8 9 M1 = 7 2 8  3 6 k + 9 1  9 2 k + ( 3 6 k + 9 2 k + 8 9 ) A1 Separates 36k + 92k +89 A1 Connects the divisibility with the is divisible by 91 because both 72836k +9192k and are divisible by 91. hypothesis. Hence, [by induction,] 3 6 n + 9 2 n + 8 9 is divisible by 91for every positive integer n. A1) Must have previous three marks and state the result in full. OR Alternative method for 2 3 6 + 9 2 + 8 9 = 9 1 0 is divisible by 91. (B1 Checks base case. Assume that ( 3 6 k + 9 2 k + 8 9 ) is divisible by 91 for some positive integer k. B1 States inductive hypothesis, making it clear it is an assumption. B0 if assume true for ALL values of k. Then 36(k+1) +92k+1+89−(36k +92k +89) M1 Considers difference between consecutive terms. = = ( 9 7 1 2 ( 8 8 )  3 3 6 k 6 k + + ( 9 9 1 2 ) k 9 ) 2 k 36k +92k +89 A1 A1 Connects the divisibility with the hypothesis. Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 2 Hence, [by induction,] © Cambridge University Press & Assessment 2026 Page 11 of 23 3 6 n + 9 2 n + 8 9 is divisible by 91 for every positive integer n. A1) Must have previous three marks and state the result in full. 6 Question Answer Marks Guidance

3  3 −3 −1 1  B1 P−1 =−1   = 2 2  6 −7 5   7 −5  6 6 © Cambridge University Press & Assessment 2026 Page 11 of 20 P D P − 1 =  5 7 3 3   1 0 0 3   − 76 12 − 12 56  =  5 7 9 9   − 76 12 − 12 56  M1 Base case, multiplies P D or D P − 1 =  − 72 12 − 12 52  . P D P − 1 =  8 7 − − 5 4  = A A1 Reference to base case Assume that it is true for n = k , so Ak =PDkP−1 . B1 States inductive hypothesis. Then Ak+1 =AkA=PDkP−1PDP−1 =PDk+1P−1 M1A1 Multiplies A k with A and substitutes A k = P D k P − 1 . So A n = P D n P − 1 is true for all positive [integers] n [by induction]. A1 States conclusion. 7 Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

3(a) © Cambridge University Press & Assessment 2026 Page 13 of 25 u 1  1 [given] B1 Base case seen. Assume that u 1 for some positive integer k. B1 States inductive hypothesis. k Accept ‘Let u k  1 be true’ or ‘Suppose u k  1 .’ EITHER: (M1 Considers u −1, and puts over k+1 u 2 +2u +2 u 2 +2u +2−(u +4) u 2 +u −2 common denominator. Numerator Then u −1= k k −1= k k k = k k k+1 u +4 u +4 u +4 need not be simplified for this mark. k k k (u −1)(u +2) = k k 0. u +4 k Or ( u k + ( u 12 k ) 2 + − 4 14) − 2  0 A1 Factorises Or completes the square. Since u 1 [by induction,] u 1 for all positive [integers] n. A1) Must have previous M1 A1 and also k n explicitly connect the assumption u k  1 with their statement. No wrong working seen. OR: Alternative for 3(a) M1 A1 A1marks Then u k + 1 − 1 = u k 2 + u k 2 u + k 4 + 2 − 1 = u k 2 + 2 u k u + k 2 + − 4 ( u k + 4 ) = u k 2 u + k u + k 4 − 2 (M1 Considers u k + 1 − 1 , and puts over common denominator. Numerator need not be simplified for this mark. u k 2  1 AND u k  1 so u 2 +u −21+1−2=0 A1 Both must be seen. k k Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 3(a) Since © Cambridge University Press & Assessment 2026 Page 14 of 25 u k  1 [by induction,] u 1 for all positive [integers] n. A1) Must have previous M1 A1and also n state u k  1 either in the conclusion or in the argument for the M1 A1. No wrong working seen. Allow for all positive integers [n]. 5 3(b) EITHER: u n + 1 − u n = u n 2 + 2 u n + u 2 n − + u 4 n ( u n + 4 ) = 2 (1 u n − + u n4 ) (M1 Considers u −u , or u −u , and n+1 n n n+1 puts over common denominator. Allow one numerical or sign error for M1. A1 So u n  1  u n + 1 − u n  0  u n + 1  u n A1) Uses u n  1 . States result. OR: Alternative for 3(b) u n + 1 − u n = − 2 + u 1 n 0 + 4 (M1 10 u −u =u −2+ −u n+1 n n u +4 n n u n  1 , u n + 4  5  u 1 n 0 + 4  2 A1 − 2 + u 1 n 0 + 4  0 u −u 0 n+1 n  u n + 1  u n A1) States result. 3 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/