RI revision programme complex plus sequences (hard)

RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 3 Revision 2 (Zeta) Topic(s): Complex Numbers and Sequences and Series (Solutions) Source of Question: 9740/2012/I/Q3(modified) 1 A sequence 1u , 2u , 3u , … is given by 1u = 2 and 1 31 6 n n uu + −= for 1n ≥ . (i) Find the exact values of 2u and 3u , and the value of 10u to 5 decimal places. [3] (ii) It is given that nul → as n →∞ . Showing your working, find the exact value of l. [2] (iii) Describe the behaviour of the sequence. [1] Solution: (i) 2 3(2) 1 5 66u −= = 3 531 16 64u  −= = From the GC, 10 0.32878 (5 dp)u = − (ii) As n →∞ , nul → and 1nul+ → We have 3( ) 1 6 ll −= 63 1ll⇒−= − 1 3l⇒= − (iii) The sequence decreases and converges to 1 3− .

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________________________________________________________ Source of Question: CJC Prelim 9758/2022/II/Q2 2 A sequence 123, , , . . .aaa is such that 123nna a nK −= −+ , where K is a constant, and 2n ≥ . (i) Given that 1 2a = and 2 4a = , find K and 3a . [2] It is known that the nth term of this sequence is given by (2 )n na p qn r= ++ , where p, q and r are constants. (ii) Find p, q and r. [3] (iii) Find 1 N n n a = ∑ in terms of N. [2] Solution: (i) Given that 1 2a = and 2 4a = , 64 2(2) 3(2) K K= ⇒− =+ 3 2 2 3(3) 2(4) 3(3) 6 5 Ka a=−+ = − += ∴ (ii) Since (2 ) ,n na p qn r= ++ 1 1 2 2 3 3 2 (2 ) (1) 2 4 (2 ) (2) 4 5 (2 ) (3) 5 a p qr a p qr a p qr = + += = + +=  = + +=  ⇒  ⇒ ⇒ Solving this system of linear equations : ∴ 1 2p = − , 3q = , and 0r = . 1 (2 ) 32 n nan = −+∴ (iii) ( 1 2) 32 n nan = −+

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________________________________________________________ ( ) ( ) 11 11 123 1 1 (2 ) 32 1 232 1 2 2 2 ... 2 3 1 2 3 ...2 1 2 (2 1) 3 (1 )2 21 2 3(1 2 ) (1 ) 2 NN n n nn NN n nn N N N an n N N N NN = = = = = −+ = −+ = − + + ++ + + + ++  − = − ++  −  = −+ + ∑∑ ∑∑ Source of Question: HCI Prelim 9740/2012/I/Q6(b) 3 Given that all terms in a geometric progression {}nu , 1 , 2 , 3 , . . .n = are positive with first term a and common ratio r, where r ≠ 1, the sums H and C are defined as follows: 12 ... nHuu u= + ++ , 12 11 1 ... n C uu u= + ++ . (i) By expressing H and C in terms of a and r, prove that 1 n H uuC = . [3] (ii) Express the product 123 ... nuuu u in the form of f( )n H C   , where f(n) is a function of n. [2] Solution: (i) Since {}nu is a GP with common ratio r, ( 1) 1 narH r −= − . 1 nu    is a GP with common ratio 1 r and thus we have 1 11(1 ) 11 1 11 nn n rarC ar r r − − −= = −− . 1 1 1 ( 1) ( 1) 11 nn n n n H a r ar r Cr r a ar uu − − −−= × −− = × = (ii) 23 1 12 1 2 3 ... ( 1) ... ( )( )( )...( ) () n n nn u u u a ar ar ar ar ar − +++ + − ⋅⋅ ⋅= = = ( 1) 2 nn nar − Since 21 1 n n H uu a rC −= = ,

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________________________________________________________ 12 ... nuu u⋅⋅ ⋅= ( 1) 2 nn nar − = 21 2() n nar − = 2 n H C   Source of Question: DHS Prelim 9740/2012/II/Q4 4 A finite sequence {}na has 50 terms and is such that 1 0.15nnaa+ = + for 1,2,3, ,49.n =  (i) Given that 50 1 99 ,aa = show that 1 0.075.a = [2] (ii) Find, without using a calculator, the value of 50 1 .n n a = ∑ [2] Another infinite sequence {}mb is such that 1 50ba= and 1 0.98m m b b − = for 2.m ≥ (iii) Determine the smallest value of k such that 25.kba< [2] (iv) Find the least value of h such that the sum of the first h terms of {}mb is more than 99% of its sum to infinity. [2] (v) If 1 0.98m m b b − = − instead, find 13 0 .m m b ∞ + = ∑ [2] Solution: (i) ( ) 50 1 11 1 99 49 99 1 0.15 0.075 (Shown)2 aa a da a = += = = (ii) ( ) 50 1 50 0.075 99 0.075 187.52n n a = = +× =∑ (iii) ( )( ) ( ) 25 1 99 0.075 0.98 (0.075) 24 0.15 35.8 least 36 k k ba k k − < × <+ > = (iv) Consider ( )1 1 1 0.98 0.991 0.98 1 0.98 0.98 0.01 227.9 least 228 h h b b h h − > −−  < > =

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________________________________________________________ (v) ( ) 13 0 147 3 99 0.075 1 0.98 3.82 (3s.f.) m m b bb b ∞ + = =+++ ×= −− = ∑  Source of Question: RI Prelim 9740/2011/I/Q6(modified) 5 The line l with equation y = 3x and the curve C with equation y = x2 are shown in the diagram above. The x-coordinates of the point Q1 on C is a1, where 0 < a1 < 3. A line parallel to the x -axis passing through the point kQ , ( 1k ≥ ) on the curve C intersects the line l at 1kP + , and another line parallel to the y -axis passing through 1kP + intersects C at 1kQ + . The sequence 12, , ,aa  is defined by the x -coordinates of the points 12, , ,QQ  respectively. (i) Show that the sequence satisfies 2 1 3 k k aa + = for all 1k ≥ . [2] It is given that the k-th term of the sequence is given by 2 1 1 3 3 k k aa − =  for all 1k ≥ . For the rest of the question, assume a1 =1. C : l : y = 3x O y x

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________________________________________________________ It is given that 1 1a = . (ii) Find 3a . [1] (iii) Find the area of the shaded rectangle in terms of 1and .kkaa + By considering the area of the region bounded by C, the lines x = 0, x =1 and the x-axis, give a geometrical reason why (ak − ak +1)ak +2 k =1 n ∑ < 1 9 for all n ≥1. [3] (iv) G iven that 21 1 1 1( )1 3 n n kk k aa − + = −= − ∑ , show that 12 1 1() 27 n kk k k aa a ++ = −<∑ for all n ≥1. [2] Solution: (i) The coordinates of the point kQ is given by ( ) 2,.kkaa Hence the coordinates of the point 1kP + is given by ( ) 2 1,.kkaa+ Since this point lies on the line y = 3x, we have 2 13kkaa += and thus 2 1 3 k k aa + = . (ii) Using 2 1 1 3 3 k k aa − =  , we have 312 3 113. 3 27a − = = (iii) Area of rectangle with base 1() kkaa +− and height 2 1ka + is 2 11( )( )kk kaa a ++− . Using the recurrence 2 1 3 k k aa + = we have 2 11 1 2( )( ) ( )(3 )kk k kk kaa a aa a++ + +−= − . So 12 1 ( )(3 ) n kk k k aa a ++ = −∑ represents the sum of the areas of rectangles under the curve y = x2 from 1na + to 1. This is clearly less than the exact area under the curve y = x2 from 0 to 1 which is given by 1 2 0 1d 3xx =∫ . Hence we have 1 2 12 11 11( )(3 ) ( )( ) . 39 nn kk k kk k kk aa a aa a+ + ++ = = − <⇒ − <∑∑ (iv) Since 12 13 3 n na − =  , the sequence is decreasing and thus, 3 1 27 naa≤= for 3.n ≥ Therefore, 1 2 13 3 1 11 1 1 1 21 () () () 1 ()27 11 11.27 3 27 n nn n kk k kk kk kk k n kk k aa a aa aa aa aa ++ + + = = = + = − − <− = − = −  = −<   ∑∑∑ ∑

Raffles Institution H2 Mathematics 2025 Year 6 ______________________________________________________________________________________________ Source of Question: ASRJC Prelim 9758/2019/I/Q11 6 (a) Find the sum of all integers between 200 and 1000 (both inclusive) that are not divisible by 7. [4] (b) Snowflakes can be constructed by starting with an equilateral triangle (Fig. 1), then repeatedly altering each line segment of the resulting polygon as follows: 1. Divide each outer line segments into three segments of equal length. 2. Add a n equilateral triangle that has the middle segment from step 1 as its base. 3. Remove the line segment that is the base of the triangle from step 2. 4. Repeat the above steps for a number of iterati