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 a