NYJC 2026 FM TP - RR with Complex Roots, APGP (Solutions)

FM Recurrence Relations with Complex Roots 1. 2023/VJC/FM Promo/Q5 (modified) [8marks, 14mins] The sequence  nv is given by 1 = 2,v 2 = 0v and 21 422n n nv v v++ − + = for 1.n The sequence  nw satisfies a second order homogeneous recurrence relation and ,nnv w k+= where k is a constant. (a) Write down a second order homogeneous recurrence relation for  nw and hence find the value of k. [2] (b) Obtain an expression for nv in a form with trigonometric function. [4] (c) Hence show that 42nv + can be expressed as  1 f( ) ,kn + where the two possible forms of f( )n are to be determined. [2] Solution: (a) (b) 21 2 2 4n n nv v v++− + = , 1 = 2,v 2 = 0v ( ) ( ) ( ) 21 21 21 1 1 2 2 21 Substitute into 2 2 4 2 2 4 2 2 4 For this equation to be homogeneous, 4. 4 2; 4 4 2 2 0 nn nn n n n n n n n n n n n n v w k v k w v v v k w k w k w w w w k k w v w v w w w ++ ++ ++ ++ +=  = − − + = − − − + − =  − + = − = = − = = − = − + = Auxiliary equation: 2 2 2 0− + = 2 4 8 1i2 −= =  ( ) ( ) 1 2 12 2 cos sin 44 112 2 2 = 22 4 4 2(0 ) 2 0 2 2sin 4 4 2 sin 4 n n n n n n nnw A B w A B A B wB B A nw nv   +  = +  =  = + + =  = + = = =   = − (b) ( ) 42 1 222 42 22 22 (4 2)4 2 sin 4 2 sin 2 142 4 2 ( 1), is odd 4 2 (1), is even 4(1 4 ), is odd 4(1 4 ), is even n n n n n n n nvn n n n n  + + + + + + += − = − +  −−= −  += − Trigo Form: Note that A and B must be real constants since the sequence generates real numbers. Cambridge tested this idea in H2 Math before under Integration Techniques. Line A stuff!

H2 Math Sequences and Series (Objectives – AQ and Interpretation Skills) 2. 2022/VJC/H2 Math Promo/Q11 [12marks, 22mins] In a laboratory experiment, a tank is transported to and fro between station A and station B. On the 1st visit to station A, 200 ml of water is added into the tank and on subsequent visits, 200 ml more water than the previous visit to station A will be added. That is, 400 ml and 600 ml of water will be added into the tank on the 2nd and 3rd visit to station A respectively. On each visit to station B, b ml of water is removed from the tank. It is given that the tank is empty at the start of the experiment and it starts from station A. (a) On the mth visit to station A, 3200 ml of water is added into the tank. Find m. [2] (b) It is given that immediately after the 16th visit to station B, there is more than 25 litres of water in the tank. Find the set of values that b can take. [3] In another laboratory experiment, a tank is transported to and fro between station A and station C. 10% of the water in the tank is removed on each visit to station C. It is given that the tank is empty at the start of the experiment, and it starts from station A. (c) Show that the amount of water in the tank immediately after the 3rd visit to station C is 1009.8 ml. [2] (d) Show that the amount of water in the tank immediately after the nth visit to station C is ( )1800 16200 1 0.9 . nn−− [4] (e) The experiment stops when the amount of water in the tank immediately after it leaves station C exceeds 10 litres. After the kth visit to station C, the experiment stopped. Find k. [1] Solution: (a) ( )200 1 (200) 3200 3200 2001 200 16 m m m + − = −−= = (b) Amount of water immediately after 16th visit to B = Total amount of water added to the tank in the 16 visits to A – Total amount of water removed from tank in the 16 visits to B ( ) ( )( )16 2 200 16 1 200 16 250002 b= + − −  Thus 27200 16 25000 16 2200 0 137.5 b b b −   (c) Visit (n) After visiting A After visiting C 1 200 0.9(200) 2 0.9(200) + 400 0.9[0.9(200) + 400] = 0.92(200) + (0.9)(400) 3 0.92(200)+(0.9)(400) +600 0.93(200)+(0.9)2(400) +(0.9)(600) 0.93(200)+(0.9)2(400)+(0.9)(600) = 1009.8 ml

(d) Total amount of water in tank after the nth visit to station C ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 1 2 2 12 1 2 1 2 2 0.9 200 0.9 400 0.9 600 ... 0.9 200( 1) 0.9 200 0.9 200 0.9 200 0 0.9 200 ... 0.9 2 t 2 0.9 20 3 0.9 20 0.9 200 <-- 0 GP wi h0 0 0 . . ( 1) 0 0 t n erms 00 0.9 2 0 9 200 ... 0 2 . 9 9 200 0 nn nnn n n n nn nn nn − − −− −− − = + + + + − + = + + + + + = + + + + + − ++ ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) 2 2 2 22 0.9 200 <-- GP with n-1 terms 0.9 200 <-- GP with n-2 terms 0.9 200 0. ... 0.9 200 ... 0.9 200 ... . 0.9 200 9 200 0.9 200 n n − − + + + + + + + + ++ +( )( )9 200 ( ) ( ) ( ) ( ) 1 2 11 0.9 1 0.9 1 0.9 1 0.90.9 200 0.9 200 ... 0.9 200 0.9 2001 0.9 1 0.9 1 0.9 1 0.9 nn −       − − − −= + + + +       − − − −        ( ) ( ) ( ) ( )( ) ( )( ) 1 2 1 1 1800 1 0.9 1 0.9 ... 1 0.9 1 0.9 1800 0.9 ... 0.9 0.9 nn nnn − − = − + − + + − + − = − + + + ( ) 1 0.91800 0.9 1 0.9 1800 16200 1 0.9 n n n n   −=−   − = − − 11(v) ( )1800 16200 1 0.9 10000nn− −  Visit (n) ( )1800 16200 1 0.9 nn−− 12 9975.36 13 11317.82 Hence, k = 13. Nice Qn from VJC. Useful skill to learn.