Summation of series

2(a) n−1n(n+1)−1n(n+1)(2n+1) M1A1 Substitutes correct formulae from MF19. 2 6 Expanding brackets correctly. 1n−n2 −1n3 A1 Simplifies by collecting terms. 3 3 3 © UCLES 2021 Page 6 of 14 2(b) r+1 r (r+1) r2 +1 −r r2 +2r+2 1−r−r2 M1 A1 Puts over a common denominator and expands, AG. − = ( )( ) = ( )( ) (r+1)2 +1 r2 +1 r2 +2r+2 r2 +1 r2 +2r+2 r2 +1 n 1−r−r2 n  r+1 r  M1 A1 Shows at least three complete terms including first and  = − ( )( )   last. Cancellation may be implicit. r=1 r2 +2r+2 r2 +1 r=1 (r+1)2 +1 r2 +1 2 1 3 2 4 3 n+1 n = − + − + − ++ − 5 2 10 5 17 10 (n+1)2 +1 n2 +1 1 n+1 A1 ISW =− + 2 ( n+1 )2 +1 5 2(c) −1 B1 FT FT from their answer to part (b). 2 1 Question Answer Marks Guidance

2(a) n−1n(n+1)−1n(n+1)(2n+1) M1A1 Substitutes correct formulae from MF19. 2 6 Expanding brackets correctly. 1n−n2 −1n3 A1 Simplifies by collecting terms. 3 3 3 © UCLES 2021 Page 6 of 14 2(b) r+1 r (r+1) r2 +1 −r r2 +2r+2 1−r−r2 M1 A1 Puts over a common denominator and expands, AG. − = ( )( ) = ( )( ) (r+1)2 +1 r2 +1 r2 +2r+2 r2 +1 r2 +2r+2 r2 +1 n 1−r−r2 n  r+1 r  M1 A1 Shows at least three complete terms including first and  = − ( )( )   last. Cancellation may be implicit. r=1 r2 +2r+2 r2 +1 r=1 (r+1)2 +1 r2 +1 2 1 3 2 4 3 n+1 n = − + − + − ++ − 5 2 10 5 17 10 (n+1)2 +1 n2 +1 1 n+1 A1 ISW =− + 2 ( n+1 )2 +1 5 2(c) −1 B1 FT FT from their answer to part (b). 2 1 Question Answer Marks Guidance

1(a) sin(r+1) sinr sin(r+1)cosr−cos(r+1)sinr sin(r+1−r) M1 Applies all relevant correct addition formulae − = = cos(r+1) cosr cos(r+1)cosr cos(r+1)cosr from MF19 for the route chosen and combines into a single fraction. sin(r+1) sinr sinrcos1+cosrsin1 sinr OR − = − cos(r+1) cosr cos(r+1) cosr sinrcosrcos1+cos2rsin1−sinrcosrcos1+sin2rsin1 = cos(r+1)cosr sin1 A1 SC B2 for this route completely correct to AG. = cos(r+1)cosr tanr+tan1 tanr+tan1−tanr+tan2rtan1 −tanr = 1−tanrtan1 1−tanrtan1 tan1sec2r tan1 sin1 = = = 1−tanrtan1 cosr(cosr−sinrtan1) cosr(cosrcos1−sinrsin1) sin1 = cosrcos(r+1) 2 1(b) n M1 A1 Shows enough terms for cancellation to be clear. [ ] sin1 u =tan2−tan1+tan3−tan2+...+tan(n+1)−tann There must be three complete terms including first r r=1 and last. Cancelling may be implied. n tan(n+1)−tan1 A1 OE and ISW. u = r sin1 r=1 3 © UCLES 2021 Page 6 of 16 1(c) tan(n+1) oscillates as n→∞ so u +u +u + does not converge. B1 States ‘oscillates’ or refers to diverging values of 1 2 3 tan(n+1), or states that tan(n+1)does not tend to a limit. 1

2(a) n ( ) M1 A1 Expands and substitutes formulae.  r3+3r2 +2r = 1n2(n+1)2 +1n(n+1)(2n+1)+n(n+1) At least two formulae used for M1. 4 2 r=1 ( ) = 1n(n+1) n2 +n+4n+2+4 = 1n(n+1)(n+2)(n+3) A1 4 4 3 © UCLES 2021 Page 6 of 14 2(b) 1 1 1 1 M1 A1 Finds partial fractions. = − + r(r+1)(r+2) 2r r+1 2(r+2) n 1 M1 A1 Shows enough terms for cancelation to be clear.  = 1 f(1)− f(2)+1 f(3) r(r+1)(r+2) 2 2 r=1 +1 f(2)− f(3)+1 f(4) 1 2 2 Where f(x)= +1 f(3)− f(4)+1 f(5) r 2 2  +1 f(n−1)− f(n)+1 f(n+1) 2 2 +1 f(n)− f(n+1)+1 f(n+2) 2 2 1 1 1 A1 ISW = − + 4 2 ( n+1 ) 2 ( n+2 ) 5 2(c) 1 B1 FT FT from their answer to part (b). 4 1 © UCLES 2021 Page 7 of 14

3(a) n B1 Separates logarithms into correct form using a ( lnr−2ln(r+1)+ln(r+2) ) difference. r=1 Or as logarithm of product. ln1−2ln2+ln3 M1 A1 Shows enough terms to make cancellation clear. ln2−2ln3+ln4 ln3−2ln4+ln5  ln(n−1)−2lnn+ln(n+1) lnn−2ln(n+1)+ln(n+2) n+2 A1 AG [ln1]−ln2−ln(n+1)+ln(n+2)=ln . 2(n+1) 4 © UCLES 2021 Page 7 of 16 3(b) S =−ln2 B1 States sum to infinity. AEF S n −S =ln    n n + +1 2   <0.01 leading to n+2<e0.01(n+1) M1 Forms inequality Least value of n is 99 A1 CAO 3 Question Answer Marks Guidance

4(a) (−2)2 −2(3) M1 Uses formula for sum of squares. −2 A1 2 4(b) α3+β3+γ3 =−2(−2)−3(−2)−3(3) M1 Uses original equation or formula for sum of cubes. 1 A1 AG 2 © UCLES 2021 Page 8 of 16 4(c) (α+r )3 =α3 +3α2r+3αr2 +r3 B1 Expands. n ( ) n ( ) M1 A1 Collects like terms and uses results from parts (a) and  (α+r )3 +(β+r )3 +(γ+r )3 = 1+3(−2)r+3(−2)r2 +3r3 (b). r=1 r=1 ( ) ( ) n−6 1n(n+1) −6 1n(n+1)(2n+1) + 3n2(n+1)2 M1 Applies formulae from MF19. 2 6 4 n−3n(n+1)−n(n+1)(2n+1)+ 3n2(n+1)2 4 n+1n(n+1) (−12−4(2n+1)+3n(n+1) ) M1 A1 Simplifies. 4 ( ) n+1n(n+1) 3n2 −5n−16 4 6 © UCLES 2021 Page 9 of 16

2(a) n ( ) M1 A1 Expands and substitutes formulae.  r3+3r2 +2r = 1n2(n+1)2 +1n(n+1)(2n+1)+n(n+1) At least two formulae used for M1. 4 2 r=1 ( ) = 1n(n+1) n2 +n+4n+2+4 = 1n(n+1)(n+2)(n+3) A1 4 4 3 © UCLES 2021 Page 6 of 14 2(b) 1 1 1 1 M1 A1 Finds partial fractions. = − + r(r+1)(r+2) 2r r+1 2(r+2) n 1 M1 A1 Shows enough terms for cancelation to be clear.  = 1 f(1)− f(2)+1 f(3) r(r+1)(r+2) 2 2 r=1 +1 f(2)− f(3)+1 f(4) 1 2 2 Where f(x)= +1 f(3)− f(4)+1 f(5) r 2 2  +1 f(n−1)− f(n)+1 f(n+1) 2 2 +1 f(n)− f(n+1)+1 f(n+2) 2 2 1 1 1 A1 ISW = − + 4 2 ( n+1 ) 2 ( n+2 ) 5 2(c) 1 B1 FT FT from their answer to part (b). 4 1 © UCLES 2021 Page 7 of 14

1(a) 1 1 1 1  M1 A1 Finds partial fractions.     (ar1)(ara1) aar1 ara1 n 1 1 1 1 1 1 1 1  M1 Writes at least three complete terms,          including first and last. (ar1)(ara1) aa1 2a1 2a1 3a1 an1 ana1 r1 n 1 1 1 1  A1 OE ISW      (ar1)(ara1) aa1 ana1 r1 4 1(b)  1 1 1 M1 A1 Finds sum to infinity, forms quadratic.    leading to a2 a60 (ar1)(ara1) a2 a 6 r1 a2 A1 CAO 3 © UCLES 2022 Page 6 of 15

1(a) 1 1 1 1  M1 A1 Finds partial fractions.     (ar1)(ara1) aar1 ara1 n 1 1 1 1 1 1 1 1  M1 Writes at least three complete terms,          including first and last. (ar1)(ara1) aa1 2a1 2a1 3a1 an1 ana1 r1 n 1 1 1 1  A1 OE ISW      (ar1)(ara1) aa1 ana1 r1 4 1(b)  1 1 1 M1 A1 Finds sum to infinity, forms quadratic.    leading to a2 a60 (ar1)(ara1) a2 a 6 r1 a2 A1 CAO 3 © UCLES 2022 Page 6 of 15

4(a) n M1 A1 Shows enough complete terms to make pattern of erx e2x 2ex 1  e3x 2e2x ex cancelling clear r1 e4x 2e3x e2x GP method. e5x 2e4x e3x  e2x 2ex 1  n erx   e2x 2ex 1  ex (ex)n 1  ex 1 r1 e(n1)x 2enx e(n1)x e(n2)x 2e(n1)x enx ex e2x e(n1)x e(n2)x A1 OE ex(ex 1)(enx 1) 3 4(b) x0 B1 Accept x0. enx 0as n leading to u u u ex e2x M1 A1 Must see enx 0[as n] agreeing with their set of 1 2 3 x 3 4(c) n n M1* Uses laws of logarithms correctly. lnu (rxln(ex 1)2) r r1 r1 n dM1 A1 n lnu  1xn(n1)nln(ex 1)2 Applies r  1n(n1). r 2 2 r1 r1 AEF for nln(ex 1)2 3 © UCLES 2022 Page 12 of 17

3(a) 8r3+12r2 +6r+1− ( 8r3−12r2 +6r−1 ) =24r2 +2 M1 Expands. n M1 A1 A1 Sums both sides and uses method of differences 24r2 +2n=(2n+1)3−1 with sufficient complete terms. r=1 n A1 AG. 24r2 =8n3+12n2 +4n=4n(n+1)(2n+1) r=1 5 © UCLES 2022 Page 7 of 15 3(b) 2n n 2n n M1 Relates with sum of squares. S =r2 +2(2r)2 =r2 +8r2 2n r=1 r=1 r=1 r=1 S = 1(2n)(2n+1)(4n+1)+ 8n(n+1)(2n+1) M1 n 2n 6 6 Applies r2 =1n(n+1)(2n+1). 6 r=1 S = 1n(2n+1)(4n+1+4n+4)= 1n(2n+1)(8n+5) A1 n 2n 3 3 r = 1n(n+1). 2 r=1 Alternative for 3(b) n n n M1 Relates with sum of squares. S =(2r−1)2 +3(2r)2 =( 4r2 −4r+1+12r2) 2n r=1 r=1 r=1 16n(n+1)(2n+1) n(n+1) M1 Applies −4 +n 6 2 n n r2 =1n(n+1)(2n+1) and r = 1n(n+1) 6 2 r=1 r=1 n( 16n2 +18n+5 ) = n (2n+1)(8n+5) A1 3 3 3 3(c) 16 B1 FT 3 1 © UCLES 2022 Page 8 of 15

1(a) n n M1 Uses List of formulae (MF19). r(r+2)=r2 +2r=1n(n+1)(2n+1)+n(n+1) 6 r=1 r=1 1n(n+1)(2n+7) A1 At least all like terms collected and 6 fractions simplified. 2 1(b) 1 11 1  M1 A1 Expresses in partial fractions. =  −  r(r+2) 2r r+2 n 2 1 1 1 M1 Writes enough terms.  =1− + − r(r+2) 3 2 4 r=1 1 1 1 1 + − + − 3 5 4 6 1 1 1 1 + − + − n−1 n+1 n n+2 n 1 13 1 1  A1 OE  = − −   r(r+2) 22 n+1 n+2 r=1 4 1(c) 3 B1 FT FT their part (b) of correct form. 4 1 © UCLES 2022 Page 6 of 15

3(a) 8r3+12r2 +6r+1− ( 8r3−12r2 +6r−1 ) =24r2 +2 M1 Expands. n M1 A1 A1 Sums both sides and uses method of differences 24r2 +2n=(2n+1)3−1 with sufficient complete terms. r=1 n A1 AG. 24r2 =8n3+12n2 +4n=4n(n+1)(2n+1) r=1 5 © UCLES 2022 Page 7 of 15 3(b) 2n n 2n n M1 Relates with sum of squares. S =r2 +2(2r)2 =r2 +8r2 2n r=1 r=1 r=1 r=1 S = 1(2n)(2n+1)(4n+1)+ 8n(n+1)(2n+1) M1 n 2n 6 6 Applies r2 =1n(n+1)(2n+1). 6 r=1 S = 1n(2n+1)(4n+1+4n+4)= 1n(2n+1)(8n+5) A1 n 2n 3 3 r = 1n(n+1). 2 r=1 Alternative for 3(b) n n n M1 Relates with sum of squares. S =(2r−1)2 +3(2r)2 =( 4r2 −4r+1+12r2) 2n r=1 r=1 r=1 16n(n+1)(2n+1) n(n+1) M1 Applies −4 +n 6 2 n n r2 =1n(n+1)(2n+1) and r = 1n(n+1) 6 2 r=1 r=1 n( 16n2 +18n+5 ) = n (2n+1)(8n+5) A1 3 3 3 3(c) 16 B1 FT 3 1 © UCLES 2022 Page 8 of 15

2(a) (4)2 2(6) M1 Uses formula for sum of squares. 4 A1 2 2(b) r2 2 2rr2 B1 Expands.  n  r2 r2 r2  n  42(4)r3r2 M1 A1 Collects like terms and uses 4 and their result from part (a). r1 r1 4n4nn1 1nn12n1 M1 Applies formulae from MF19. 2 4n2  1nn12n1 M1 A1 Simplifies. 2 n  4n 1  2n2 3n1  2 n  n2 5n 1  2 2 6 © UCLES 2023 Page 7 of 16

3(a) 1 1 1 1  M1 A1 Finds partial fractions.     (kr1)(krk1) kk(r1)1 kr1 n 1 1 1 1 1 1 1  M1 Writes at least three complete terms, including   1      last. (kr1)(krk1) k k1 k1 2k1 k(n1)1 kn1 r1 1 1  A1 n 1  OE e.g. k kn1 kn1 4 3(b) 1 B1 k 1 3(c) n2 1 n2 1 n1 1 M1 Or applies the method of differences again.    (kr1)(krk1) (kr1)(krk1) (kr1)(krk1) rn r1 r1 1 1  1  1 1 1  A1 n2 (n1) k  1 kn21    1 k(n1)1    k  k(n1)1  kn21  OE e.g. kn2 1  k(n1)1 2 © UCLES 2023 Page 8 of 16

2(a) (4)2 2(6) M1 Uses formula for sum of squares. 4 A1 2 2(b) r2 2 2rr2 B1 Expands.  n  r2 r2 r2  n  42(4)r3r2 M1 A1 Collects like terms and uses 4 and their result from part (a). r1 r1 4n4nn1 1nn12n1 M1 Applies formulae from MF19. 2 4n2  1nn12n1 M1 A1 Simplifies. 2 n  4n 1  2n2 3n1  2 n  n2 5n 1  2 2 6 © UCLES 2023 Page 7 of 16

3(a) 1 1 1 1  M1 A1 Finds partial fractions.     (kr1)(krk1) kk(r1)1 kr1 n 1 1 1 1 1 1 1  M1 Writes at least three complete terms, including   1      last. (kr1)(krk1) k k1 k1 2k1 k(n1)1 kn1 r1 1 1  A1 n 1  OE e.g. k kn1 kn1 4 3(b) 1 B1 k 1 3(c) n2 1 n2 1 n1 1 M1 Or applies the method of differences again.    (kr1)(krk1) (kr1)(krk1) (kr1)(krk1) rn r1 r1 1 1  1  1 1 1  A1 n2 (n1) k  1 kn21    1 k(n1)1    k  k(n1)1  kn21  OE e.g. kn2 1  k(n1)1 2 © UCLES 2023 Page 8 of 16

2(a) 1  1  M1 Substitutes formulae for r2 and r. 6 nn12n1 6 nn1  5n 6  2  2n36n2n A1 2 © UCLES 2023 Page 6 of 16 2(b) 6r2 6r5 5 B1 Divides by denominator. 6 r2 r r(r1) 1 1 1 B1 Finds partial fractions.   rr1 r r1 n 1 1 1 1 1 1 M1 Writes at least three complete terms, including the last  1     term, to show cancelation. rr1 2 2 3 n n1 r1 n 6r2 6r5 5 A1 6n2 n  6n5 OE e.g. r2 r n1 n1 r1 4 2(c) 2n 6r2 6r5 n 6r2 6r5 M1 Or uses method of differences again.   r2 r r2 r r1 r1 5 5 12n5 6n5 2n1 n1 5 5 5 5 A1 5n 12n5 6n5 6n  Or 6n . OE, like terms collected. 2n1 n1 2n1 n1 n12n1 2 © UCLES 2023 Page 7 of 16

1(a) r2 +2r+1−r2 =2r+1 B1 Expands n M1 A1 Uses method of differences and sums both sides. 2r+n=(n+1)2 −12 r=1 n A1 AG. 2r=n2 +n=n(n+1) r=1 4 1(b) n n M1 Relates with r. (r+a)=r+an r=1 r=1 1n(n+1)+an=n M1 n 2 Applies r= 1n(n+1). 2 r=1 a= 1(1−n) A1 2 3 © UCLES 2023 Page 6 of 15

1(a) 1n(n+1)(2n+1)+ 3n(n+1)+n M1 A1 Substitutes correct formulae from 2 2 MF19. n3+3n2+3n A1 Simplifies 3 1(b) 1 1 (r+1)3−r3 r3+3r2 +3r+1−r3 3r2 +3r+1 M1 A1 Puts over a common denominator and − = = = expands, AG. r3 (r+1)3 r3(r+1)3 r3(r+1)3 r3(r+1)3 n 3r2 +3r+1 n  1 1  M1 A1 Shows three complete terms, including  =  −  last. r3(r+1)3 r3 (r+1)3  r=1 r=1 1 1 1 1 1 =1− + − + + − 23 23 33 n3 (n+1)3 1 A1 1− (n+1)3 5 1(c) 1 B1FT FT from their answer to part (b). 1 © UCLES 2023 Page 6 of 15

1(a) r2 +2r+1−r2 =2r+1 B1 Expands n M1 A1 Uses method of differences and sums both sides. 2r+n=(n+1)2 −12 r=1 n A1 AG. 2r=n2 +n=n(n+1) r=1 4 1(b) n n M1 Relates with r. (r+a)=r+an r=1 r=1 1n(n+1)+an=n M1 n 2 Applies r= 1n(n+1). 2 r=1 a= 1(1−n) A1 2 3 © UCLES 2023 Page 6 of 15

3(a) N N N N B1 Expands. r(r1)3r43r3 7r2 4r r1 r1 r1 r1 3N2N 12  7 N(N 1)(2N 1)2NN 1 M1 Substitutes formulae from MF19. 4 6 1 N  9N  N2 2N 1  14  2N2 3N 1  24N 24  A1 12 1 N  9N346N2 75N 38   1 N(N 1)(N 2)(9N 19) AG 12 12 3 © Cambridge University Press & Assessment 2024 Page 7 of 16 3(b) 3r4 4 1 M1 A1 Finds partial fractions.   r(r1) r r1 N 3r4 1 r1  1 4 1  1 4 1  1  4 1  M1 Writes at least three terms, including last.                   r(r1)4 42 1 2 43 2 3 4N1N N 1 r1 1 1 A1  4 4N1(N 1) 4 3(c) 1 B1 CAO 4 1 Question Answer Marks Guidance

3(a) N N N N B1 Expands. r(r1)3r43r3 7r2 4r r1 r1 r1 r1 3N2N 12  7N(N 1)(2N 1)2NN 1 M1 Substitutes formulae from MF19. 4 6 1 N  9N  N2 2N 1  14  2N2 3N 1  24N 24  A1 12 1 N  9N3 46N2 75N 38   1 N(N 1)(N 2)(9N 19) AG 12 12 3 © Cambridge University Press & Assessment 2024 Page 7 of 16 3(b) 3r4 4 1 M1 A1 Finds partial fractions.   r(r1) r r1 N 3r4 1 r1  1 4 1  1 4 1  1  4 1  M1 Writes at least three terms, including last.                   r(r1)4 42 1 2 43 2 3 4N1N N 1 r1 1 1 A1  4 4N1(N 1) 4 3(c) 1 B1 CAO 4 1 Question Answer Marks Guidance

4(a) 12  1(1)(2)(3) so H is true. B1 Checks base case. 6 1 k B1 States inductive hypothesis. Assume that r2  1k(k1)(2k1). 6 r1 k1 M1 Considers sum to k1. r2  1k(k1)(2k1)k12 6 r1 1(k1)  2k2 k6k6   1(k1)  2k2 7k6   1(k1)k22k3 A1 6 6 6 So H is true. By induction, H is true for all positive integers n. A1 States conclusion. k1 n 5 4(b) 35 15 55 35 75 552n15 (2n1)5 M1 Applies method of differences to LHS, writes complete terms for at least three values of r including the first and last. 2n15 1 A1 SCB1 for insufficient complete terms shown. 160S  40n(n1)(2n1)2n M1 A1 Sums RHS and applies standard formulae. n 3 160S 2n15  40n(n1)(2n1)(2n1) M1 Makes S n the subject and factorises. n 3 160S (2n1)  2n14  40n(n1)1  n 3 2n1 16n4 32n3  32n2 16n  16n2n1 3n3 6n2 2n1  A1 AG 3 3 3 S  1 n(n1)(2n1)(3n2 3n1) n 30 6 © Cambridge University Press & Assessment 2024 Page 10 of 16 4(c) n5S  1 1 12 1 3 3  1  M1 Divides by n5. n 30 n n n n2 1 A1 5 2 Question Answer Marks Guidance

4(a) 5k  1 1  M1 A1 Finds partial fractions. (Don’t allow a =k  −  substitution of a value for k.) (5r+k)(5r+5+k) 5r+k 5r+5+k n 5k  1 1 1 1 1 1  M1 Writes at least three terms, including last.  =k  − + − + + −  (Allow any value of k.) (5r+k)(5r+5+k) 5+k 10+k 10+k 15+k 5n+k 5n+5+k r=1  1 1  A1 =k  −  5+k 5n+5+k 4 4(b) k 1 5 M1 A1 = 3k =5+kk = 5+k 3 2 2 4(c) n2 5k n2 5k n−1 5k M1 Or applies the method of differences again.  = − (5r+k)(5r+5+k) (5r+k)(5r+5+k) (5r+k)(5r+5+k) r=n r=1 r=1  1 1   1 1   1 1  A1FT FT on their value of k (must be substituted in). =k − −k − =k −  5+k 5n2 +5+k 5+k 5n+k 5n+k 5n2 +5+k 1 1 = − 2n+1 2n2 +3 2 © Cambridge University Press & Assessment 2024 Page 9 of 15

5(a) xf(1) −xf(2) +xf(2) −xf(3) + +xf(n) −xf(n+1) M1 Writes at least three terms, including last. = xf(1) −xf(n+1) A1 2 © Cambridge University Press & Assessment 2024 Page 9 of 16 5(b) 0x1 B1 Accept 0x1 n u =1−xln(n+1) r r=1  B1 Without wrong working. u =1 [for x1] r r=1  B1 u =0 for x=1 r r=1 3 5(c) Uses x2log x r =r2 M1 S =1−(n+1)2 =−n2−2n A1 n N M1 Substitutes formulae from MF19. −n2 −2n=−1 N(N+1)(2N+1)−N(N+1) 6 n=1 −1 N(N +1)(2N +7) A1 6 4 © Cambridge University Press & Assessment 2024 Page 10 of 16

4(a) 5k  1 1  M1 A1 Finds partial fractions. (Don’t allow a =k  −  substitution of a value for k.) (5r+k)(5r+5+k) 5r+k 5r+5+k n 5k  1 1 1 1 1 1  M1 Writes at least three terms, including last.  =k  − + − + + −  (Allow any value of k.) (5r+k)(5r+5+k) 5+k 10+k 10+k 15+k 5n+k 5n+5+k r=1  1 1  A1 =k −  5+k 5n+5+k 4 4(b) k 1 5 M1 A1 = 3k =5+kk = 5+k 3 2 2 4(c) n2 5k n2 5k n−1 5k M1 Or applies the method of differences again.  = − (5r+k)(5r+5+k) (5r+k)(5r+5+k) (5r+k)(5r+5+k) r=n r=1 r=1  1 1   1 1   1 1  A1FT FT on their value of k (must be substituted in). =k − −k − =k −  5+k 5n2 +5+k 5+k 5n+k 5n+k 5n2 +5+k 1 1 = − 2n+1 2n2 +3 2 © Cambridge University Press & Assessment 2024 Page 9 of 15

1(a) (2−3r)(5−3r)=9r2 −21r+10 B1 9 (1n(n+1)(2n+1) ) −21 (1n(n+1) ) +10n M1 Substitutes formulae for r2 and r. 6 2 =3n3−6n2 +n A1 3 1(b) 1 1 1 1  B1 Finds partial fractions. =  −  (2−3r)(5−3r) 32−3r 5−3r n 1 1 1 1 1 1 1  1 1 1  M1 A1 Writes at least three terms, including first and  =  −  +  −  +  −  last, to demonstrate cancelation. (2−3r)(5−3r) 3−1 2 3−4 −1 3−7 −4 r=1 1 1 1  + +  −  32−3n 5−3n 1 1 A1 =− + 6 3(2−3n) 4 1(c)  1 1 B1 Allow FT from (b) if of similar form.  =− (2−3r)(5−3r) 6 r=1 1 © Cambridge University Press & Assessment 2025 Page 9 of 18

1(a) (2−3r)(5−3r)=9r2 −21r+10 B1 9 (1n(n+1)(2n+1) ) −21 (1n(n+1) ) +10n M1 Substitutes formulae for r2 and r. 6 2 =3n3−6n2 +n A1 3 1(b) 1 1 1 1  B1 Finds partial fractions. =  −  (2−3r)(5−3r) 32−3r 5−3r n 1 1 1 1 1 1 1  1 1 1  M1 A1 Writes at least three terms, including first and  =  −  +  −  +  −  last, to demonstrate cancelation. (2−3r)(5−3r) 3−1 2 3−4 −1 3−7 −4 r=1 1 1 1  + +  −  32−3n 5−3n 1 1 A1 =− + 6 3(2−3n) 4 1(c)  1 1 B1 Allow FT from (b) if of similar form.  =− (2−3r)(5−3r) 6 r=1 1 © Cambridge University Press & Assessment 2025 Page 9 of 18

4(a) w −w =(r+1)(r+2) (r+10)−r(r+1)(r+2) (r+9) M1 Correct expression seen for the difference. r+1 r (r+1)(r+2) (r+9)(r+10−r)=10(r+1)(r+2) (r+9) A1 Must see the (r + 10 – r) AG. 2 4(b) 1 (w −w )+ 1 (w −w )+ + 1 (w −w ) M1 A1 Writes at least three terms, including last. Can 10 2 1 10 3 2 10 n+1 n be implied by correct final answer. = 1 (w −w )= 1 (n+1)(n+2) (n+10)−10!= 1 (n+1)(n+2) (n+10)−9! A1 ISW 10 n+1 1 10 10 10 3 © Cambridge University Press & Assessment 2025 Page 12 of 19 4(c) v +v + +v = xw 2 −xw 1 +xw 3 −xw 2 + +xw n+1 −xw n = xw n+1 −xw 1 M1 1 2 n For −1x1,sum to infinity is −xw 1 =−x10! A1 For x=1, sum to infinity is 0. A1 3 Question Answer Marks Guidance

1(a) n n(n+1) M1 Uses formula for r. 2r+n=2 +n 2 r=1 =n(n+2) A1 Or n2 +2n 2 1(b) 1 1 r2 +2r+2−r2 −1 2r+1 B1 Puts over common denominator. − = = r2 +1 r2 +2r+2 ( r2 +1 )( r2 +2r+2 ) ( r2 +1 )( r2 +2r+2 ) 1 1(c) 1 1 1 1 1 1 1 1 M1 Shows enough complete terms to demonstrate − + − + − + + − 2 5 5 10 10 17 n2 +1 n2 +2n+2 cancellation. 1 1 A1 − 2 n2 +2n+2 2 1(d) 1 B1 FT 2 1 © Cambridge University Press & Assessment 2025 Page 9 of 18

1(a) 2n2(n+1)2 +2n(n+1)(2n+1)+2n(n+1)+3n M1 A1 Substitutes correct formulae from MF19. =2n4 +8n3+10n2 +7n A1 Simplifies. 3 1(b) 1 1 (r+1)4 −r4 r4 +4r3+6r2 +4r+1−r4 4r3+6r2 +4r+1 M1 A1 Puts over a common denominator and − = = = expands, AG. r4 (r+1)4 r4(r+1)4 r4(r+1)4 r4(r+1)4 n 4r3 +6r2 +4r+1 n  1 1  M1 A1 Shows three complete terms, including  =  −  first and last. r4(r+1)4 r4 (r+1)4  r=1 r=1 1 1 1 1 1 =1− + − + + − 24 24 34 n4 (n+1)4 1 A1 =1− (n+1)4 5 1(c) 1 B1FT FT from their answer to part (b). 1 Question Answer Marks Guidance

1(a) n M1 A1 Substitutes formulae. For the 1st A mark ( r3−r ) =1n2(n+1)2 −1n(n+1) allow any letter for n. 4 2 r=1 = 1n(n+1)(n−1)(n+2) A1 Factorises. 4 3 1(b) r+3 2 3 1 M1 A1 Finds partial fractions. M mark is for an = − + r3−r r−1 r r+1 equation that would lead to values for A, B and C. n r+3 M1 A1 Shows five complete terms including  = 2f(1)−3f(2)+ f(3) first two and last two. (r−1)r(r+1) r=2 +2f(2)−3f(3)+ f(4) +2f(3)−3f(4)+ f(5)  +2f(n−2)−3f(n−1)+ f(n) +2f(n−1)−3f(n)+ f(n+1) 3 2 1 A1 SCB1 if insufficient terms to show all = − + of the cancelling, but must start at r = 2 2 n n+1 5 1(c) 3 B1FT FT from their part (b). Needs at least 2 two terms in part (b) with one involving a number and a term in n. 1 © Cambridge University Press & Assessment 2025 Page 9 of 19

1(a) 2n2(n+1)2 +2n(n+1)(2n+1)+2n(n+1)+3n M1 A1 Substitutes correct formulae from MF19. =2n4 +8n3+10n2 +7n A1 Simplifies. 3 1(b) 1 1 (r+1)4 −r4 r4 +4r3+6r2 +4r+1−r4 4r3+6r2 +4r+1 M1 A1 Puts over a common denominator and − = = = expands, AG. r4 (r+1)4 r4(r+1)4 r4(r+1)4 r4(r+1)4 n 4r3 +6r2 +4r+1 n  1 1  M1 A1 Shows three complete terms, including  =  −  first and last. r4(r+1)4 r4 (r+1)4  r=1 r=1 1 1 1 1 1 =1− + − + + − 24 24 34 n4 (n+1)4 1 A1 =1− (n+1)4 5 1(c) 1 B1FT FT from their answer to part (b). 1 Question Answer Marks Guidance

3(a) n M1 A1 Expands and substitutes formulae. ( r3+6r2 +11r+6 ) = 1n2(n+1)2 +n(n+1)(2n+1)+11n(n+1)+6n 4 2 r=1 = 1n ( n3+10n2 +35n+50 ) A1 4 3 3(b) 2 1 2 1 M1 A1 Finds partial fractions. = − + (r+1)(r+2)(r+3) r+1 r+2 r+3 n 2 M1 A1 Shows enough complete terms for cancellation to be  = f(1)−2f(2)+ f(3) clear and any terms that form part of the final answer. (r+1)(r+2)(r+3) r=1 1 +f(2)−2f(3)+ f(4) f(r)= . r+1 +f(3)−2f(4)+ f(5)  +f(n−1)−2f(n)+ f(n+1) +f(n)−2f(n+1)+ f(n+2) 1 1 1 A1 1 1 = − + Accept − 6 n+2 n+3 6 (n+2)(n+3) 5 3(c) 1 B1 FT their result of similar form 6 1 © Cambridge University Press & Assessment 2025 Page 11 of 20

1(a) © Cambridge University Press & Assessment 2026 Page 9 of 23 ( 4 r − 3 ) (1 2 r + 3 ) = 4 8 r 2 − 2 4 r − 9 B1 4 8 ( 16 n ( n + 1 ) ( 2 n + 1 ) ) − 2 4 ( 12 n ( n + 1 ) ) − 9 n M1 Substitutes formulae for  r 2 and  r . =16n3+12n2 −13n A1 CAO Must be in this form. 3 1(b) 1 1  1 1  B1 Finds partial fractions. Accept equivalent =  −  form. (4r−3)(12r+3) 124r−3 4r+1 r n = 1 ( 4 r − 3 ) 1 (1 2 r + 3 ) = 1 1 2  1 1 − 1 5  + 1 1 2  1 5 − 1 9  + 1 1 2  1 9 − 1 1 3  M1 Write enough correct sections of terms so that all that are left in the answer appear plus one pair that cancel. +  + 1 1 2  4 n 1 − 3 − 4 n 1 + 1  A1 All necessary complete terms to demonstrate the cancellation. Either two at the beginning and one at the end or one at the beginning and two at the end. = 1 1 2 − 1 2 ( 4 1 n + 1 ) B1 Accept any equivalent form. 4 1(c)  r = 1 ( 4 r − 3 ) 1 (1 2 r + 3 ) = 1 1 2 B1 FT their answer for 1(b) of similar form. 1 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/

1(a) © Cambridge University Press & Assessment 2026 Page 9 of 23 ( 4 r − 3 ) (1 2 r + 3 ) = 4 8 r 2 − 2 4 r − 9 B1 4 8 ( 16 n ( n + 1 ) ( 2 n + 1 ) ) − 2 4 ( 12 n ( n + 1 ) ) − 9 n M1 Substitutes formulae for  r 2 and  r . =16n3+12n2 −13n A1 CAO Must be in this form. 3 1(b) 1 1  1 1  B1 Finds partial fractions. Accept equivalent =  −  form. (4r−3)(12r+3) 124r−3 4r+1 r n = 1 ( 4 r − 3 ) 1 (1 2 r + 3 ) = 1 1 2  1 1 − 1 5  + 1 1 2  1 5 − 1 9  + 1 1 2  1 9 − 1 1 3  M1 Write enough correct sections of terms so that all that are left in the answer appear plus one pair that cancel. +  + 1 1 2  4 n 1 − 3 − 4 n 1 + 1  A1 All necessary complete terms to demonstrate the cancellation. Either two at the beginning and one at the end or one at the beginning and two at the end. = 1 1 2 − 1 2 ( 4 1 n + 1 ) B1 Accept any equivalent form. 4 1(c)  r = 1 ( 4 r − 3 ) 1 (1 2 r + 3 ) = 1 1 2 B1 FT their answer for 1(b) of similar form. 1 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/

4(a) 16r4 +32r3+24r2 +8r+1− ( 16r4 −32r3+24r2 −8r+1 ) =64r3+16r M1 Some evidence of expansion or evidence of using difference of two squares. Allow one error overall. A1 Correct answer with like terms collected. © Cambridge University Press & Assessment 2026 Page 12 of 20 r = r = ... r =  6  4 4 1 3 4 2 5 ( n 2 n n  3 r + r = 1 − 1 − 3 + 1 6 4 4 ) 1 n r = 4 1 − r ( = 2  n ( − 2 1 n ) + 4 1 ) 4 − 1 M1 Write enough correct rows of terms so that all that are left in the answer appear plus one pair that cancel. A1 All necessary rows of terms to demonstrate the cancellation. Either two at the beginning and one at the end or one at the beginning and two at the end.  6 4 r n = 1 r 3 = 1 6 n 4 + 3 2 n 3 + 2 4 n 2 + 8 n − 8 n ( n + 1 ) M1 Applies r n = 1 r = 12 n ( n + 1 ) . n n A1 AG. Must see at least one intermediate step of 64r3 =16n2( n2 +2n+1 ) r3 =1n2(n+1)2 working before the answer given. 4 r=1 r=1 6 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/ 4(b) EITHER: © Cambridge University Press & Assessment 2026 Page 13 of 20  S 2 n  = 2 n r = 1 r 3 − 2 r n = 1 ( 2 r ) 3 (*M1 Relates with sum of cubes.  S 2 n  = 14 ( 2 n ) 2 ( 2 n + 1 ) 2 − 1 64 n 2 ( n + 1 ) 2 DM1 Applies r n = 1 r 3 = 14 n 2 ( n + 1 ) 2 . S 2 n = n 2 ( ( 2 n + 1 ) 2 − 4 ( n + 1 ) 2 ) = n 2 ( − 4 n − 3 ) A1) Accept − n 2 ( 4 n + 3 ) OR: Alternative method  S 2 n  = ( 1 3 + 3 3 + ... + ( 2 n − 1 ) 3 ) − ( 2 3 + 4 3 + ... + ( 2 n ) 3 ) (*M1 Relates with difference of cubes  S 2 n  = − 1 2 r n = 1 r 2 + 6 r n = 1 r − n  S 2 n  = − 1 6 2 n ( n + 1 ) ( 2 n + 1 ) + 6 2 n ( n + 1 ) − n DM1 Applies r n = 1 r 2 = n 6 ( n + 1 ) ( 2 n + 1 ) a n d r n = 1 r = n 2 ( n + 1 ) S =−4n3−6n2−2n+3n2+3n−n=n2(−4n−3) A1) Accept −n2(4n+3) 2n Available marks 3 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/

1(a) 1n(n+1)(2n+1)−n(n+1)+n M1 Substitutes all three correct formulae 6 (two from MF19). © Cambridge University Press & Assessment 2026 Page 9 of 25 13 n 3 − 12 n 2 + 16 n A1 Must be in this form. 2 1(b) EITHER: 1 1 1 1 − = − r2 −2r+1 r2 +2r+1 (r−1)2 (r+1)2 1 1 1 1 1 1 1 1 1 1 1 1 (M1 Writes enough correct sections of terms − + − + − + − ++ − + − so that three of the four sections which 12 32 22 42 32 52 42 62 (n−2)2 n2 (n−1)2 (n+1)2 appear in the final answer are seen. A1 All four correct sections correct and at least one cancellation pair included. A1 At least five complete n- value terms, either first three and last two, or first two and last three. 5 4 − 1 n 2 − ( n 1 + 1 ) 2 B1) OE WWW 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/ 1(b) OR: Alternative method for 1(b) © Cambridge University Press & Assessment 2026 Page 10 of 25  1 2 1 + 1 2 2 + 1 3 2 + 1 4 2 +  + ( n 1 − 2 ) 2 + ( n 1 − 1 ) 2  −  1 3 2 + 1 4 2 + 1 5 2 + 1 6 2 +  + 1 n 2 + ( n 1 + 1 ) 2  (M1 Writes enough correct sections of terms so that three of the four sections which appear in the final answer are seen. A1 All four correct sections correct and at least one cancellation pair included. A1 At least five complete n- value terms, either first three and last two, or first two and last three. WWW 5 1 1 B1) OE WWW − − 4 n2 (n+1)2 4 1(c) 5 B1 FT FT their 1(b) of similar form. 4 1 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/