SAJC Revision Worksheet on Sequences & Series

1 Revision Worksheet on Sequences and Series A. (Sigma) notation Given n r r m u   , First term = mu Last term = nu No. of terms = 1n m  B. Basic properties/rules of summation No. Property Express the following examples in a similar form to the property given (the final answer is not required) 1 ( 1) n r m k n m k     , where m n Note: r is the index and k is treated as a constant 1n m  is the number of terms in the series 10 3 4 8 4 r   (Note : can use GC to check the answer if the limits are finite known values) 2 1 1 n n r r r r ku k u     8 8 8 1 1 1 5 1 1 5 5 r r r r r r                       3   1 1 1 n n n r r r r r r r u v u v           1 1 1 2 2 n n n r r r r r r r         4 1 1 1 where n n m r r r r m r r u u u m n          13 13 4 5 1 1 ! ! ! r r r r r r        5 Given the formula for nS , i.e. 1 n r r u   , the thn term of the series, 1 1 1 1 n n n r r r r n n u u u S S           Given 36 2 n nS   , 1 1 1 3 36 6 2 2 3 3 2 2 6 3 3 2 2 2 n n n n n n n n n n u S S                        r: index of sum m: lower limit n: upper limit Note: Lower limit of a sum does not always need to be 1.

2 C. Formula for some standard series No. Formula Example Remark 1   1 1 2 n r n nr   20 1 4 r r   20 1 4 r r    (20)(21)4 2  840 Using result of 1 1 n n r r r r ku k u     2   2 1 1 2 1 6 n r n n nr    Formula will be given in the question when required   1 (3 1) n r r    2 1 3 n r r r    2 1 1 3 n n r r r r      ( 1)(2 1) ( 1)3 6 2 n n n n n        ( 1) (2 1) 1 2 n n n   2( 1)n n  Using result of 1 1 n n r r r r ku k u     3 2 3 1 ( 1) 2 n r n nr      Formula will be given in the question when required 15 3 6 ( 2)r  15 15 3 6 6 2 r r r      15 5 15 3 3 1 1 6 2 r r r r r            2 2 (15)(16) (5)(6) (15 6 1)(2)2 2               14400 225 20   14195 change lower limit to 1 D. Limit and Convergence of Series Let 1 2 3 1 ... n n r n r S u u u u u        . When we write lim nn S S   , where S is a unique finite value , it means that nS approaches the finite value S as n   .  If lim nn S  = S, where S is a finite value, the series nS converges.  If lim nn S  does not exist or is infinite, the series nS diverges.

3 E. Recurrence Relations Recurrence relations are when the nth term is expressed in terms of the previous terms. Eg. 1 3n nu u  ; 1 3; u  . To find the terms, we do the following: When n =1, 2 1 3 3(3) 9u u   When n =2, 3 2 3 3(9) 27u u   When n =3, 4 3 3 3(27) 81u u   …and so on. Do remember how to use the graphing calculator to find the nth term of a recurrence relation. Practice Problems 1 HCI JC 1 Promo 9758/2024/Q2 (a) Given that    2 1 1 2 16 n r nr n n     , find 1 1 1 2 3 rn r r r                  , in terms of n. [4] (b) Explain why 1 1 1 12 3 rn r r r                   for all positive integers n. [1] 2 NYJC JC2 Prelim 9758/2019/01/Q10 For this question, you may use the results   2 1 1 2 1 6 n r n n nr    and  3 22 1 4 1 n r n nr   . (i) Find   2 1 12 n r r r   in terms of n. [2] (ii) Find   2 1 1 n r r r   in terms of n. Hence find   1 2 2 1 n r rr    in terms of n. [5] (iii) Without using a graphing calculator, find the sum of the series      4 25 5 36 6(49) 7(64) 59 3600      . [3]

4 5 CJC JC1 Promo 9758/2024/Q3 It is given that  1 1 1 11 1 n r r r n    . (a) Find   1 5 1 1 n r r r    . [3] (b) Give a reason why the series in part (a) is convergent and state the value of  5 1 1r r r    . [2] 3 SAJC JC2 Prelim 9758/2019/02/Q4 (b) (b) (i) Cauchy’s root test states that a series of the form 0 r r a    (where 0ra  for all r) converges when lim 1,n nn a   and diverges when lim 1n nn a   . When lim 1,n nn a   the test is inconclusive. Using the test and given that lim 1n p n n   for all positive p, explain why the series 0 2 3 r x r r r   converges for all positive values of x. [3] (ii) By considering 2 2 3(1 ) 1 2 3 4y y y y       , evaluate 0 2 3 r x r r r   for the case when 1x  . [2] 4 TJC JC2 Prelim 9758/2019/02/Q2 (modified) It is given that   1 1 1 1 6 7 1 7 7 1 7 1 r NN r r r r N                   (i) Give a reason why the series   1 1 1 6 7 7 1 r r r r r              converges, and write down its value. [2] (ii) Use your answer in part (i) to find    1 1 1 6 13 7 1 2 rN r r r r               . [3]

5 6. CJC JC1 Promo 9758/2024/Q8 (b) (b) A sequence 0 1 2, , , u u u  is given by 0 400u  and 11.01n nu u x   for 1n , where x is an integer. (i) Show that    1.01 400 100 1.01 1n n nu x   . [3] (ii) Given that 16x and 1.01 ky u , where 0 16y  , find the value of k and y . [4] 7 DHS JC 1 Promo 9758/2024/Q3(a) Given that 2 1 ( 1)(2 1),6 n r nr n n     evaluate ( 1)( 3) n r n r r    in terms of n. [4] 8 ACJC JC 1 Promo 9758/2024/Q8 (a) It is given that 2 1 1 4 1 2 1 n r n r n   . (i) Explain why the series 2 1 1 4 1r r    converges and write down the value to which it converges. [2] (ii) Find  6 1 2 1 (2 3) N r r r   in terms of N, express your answer in a single fraction. [3] (b) The sequence 1 2 3, , ,u u u  is defined by 1 2u  , 1 11 , 1n n u n u     . Find the value of 2u , 3u and 4u . Hence find the value of 50 1 r r u   . [4]

6 9 RI JC 1 Promo 9758/2024/Q5 (a) Find   0 ( 2) , n r n r n    giving your answer in terms of n. [3] [You may use the result   23 2 1 1 14 n r r n n    for the rest of this question.] (b) By writing down the first two and the last two terms in the series, find   3 1 2 , n r r   giving your answer in terms of n. [3] (c) Find     3 33 3 3 3 3 31 2 3 4 5 6 2 1 2 , n n         giving your answer in terms of n. [3] End Answers 1. (a)   211 1 2 3 n n n      2. (i)    21 1 3 16   n n n n (ii)    1 1 3 2 112   n n n n ;    1 1 1 3 2 412    n n n n (iii) -108836 3. (b)(ii) 6 4. 1 7 ; 21 7 14 2 N N     5. (a) 1 1 5 2 n  (b) 1 5 6. (b)(ii) 28k  ; 14.6y 7. a) 22 1 ( 9)3 n n n   8 (ai) 1 2 (aii) 5 13(2 3) N N   (b) 26.5 9 (a)   1 42 n n n  ; (b)     2 21 2 3 94 n n   ;(c)   2 4 3n n 