Sequence of Numbers with Three Alternate Common Differences and ...

International Journal of Applied Mathematical Research, 1 (3) (2012) 259-267 c °Science Publishing Corporation www.sciencepubco.com/index.php/IJAMR

Sequence of Numbers with Three Alternate Common Differences and Common Ratios Julius Fergy T. Rabago Faculty, Department of Mathematics and Physics College of Arts and Sciences, Central Luzon State University Science City of Mu˜ noz 3120, Nueva Ecija, Philippines Email: julius [email protected] Abstract This paper talks about two types of special sequences. The first is the arithmetic sequence of numbers with three alternate common differences; and the other, is the geometric sequence of numbers with three alternate common ratios. The formulas for the general term an and the sum of the first n terms, denoted by Sn , are given respectively.

Keywords: Sequence of numbers with three alternate common differences, sequence of numbers with three alternate common ratios, general term an , the sum of the first n terms denoted by Sn .

1

Arithmetic sequence of numbers with three alternate common differences

Definition 1.1. A sequence of numbers {an } is called a sequence of numbers with three alternating common differences if the following conditions are satisfied: (i) for all k ∈ N , a3k−1 − a3k−2 = d1 , (ii) for all k ∈ N , a3k − a3k−1 = d2 , (iii) for all k ∈ N , a3k+1 − a3k = d3 , here d1 (d2 , and d3 ) is called the first (the second and the third) common differences of {an }.

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Julius Fergy T. Rabago

Example 1.2. The number sequence 1, 2, 4, 7, 8, 10, 13, 14, 16, . . . is a sequence of numbers with three alternate common differences, where d1 = 1, d2 = 2, and d3 = 3. Obviously, {an } has the following form a1 , a1 + d1 , a1 + d1 + d2 , a1 + d1 + d2 + d3 , a1 + 2d1 + d2 + d3 , a1 +2d1 +2d2 + d3 , a1 + 2d1 + 2d2 + 2d3 , a1 + 3d1 + 2d2 + 2d3 , a1 + 3d1 + 3d2 + 2d3 , . . . Theorem 1.3. The formula of the general term of an is ¹ º ¹ º jnk n−1 n+1 d1 + d2 + d3 an = a1 + 3 3 3

(1)

Proof. We prove this theorem by induction on n. Obviously, (1) holds for n = 1, 2, 3 and 4. Suppose (1) holds when n = k, hence ¹ º ¹ º ¹ º k+1 k k−1 ak = a1 + d1 + d2 + d3 3 3 3 We need to show that P(k + 1) also holds for any k ∈ N . (i.) If k = 3m − 2 , where m ∈ N , then ak+1 = ak + d1 ¹ º ¹ º ¹ º k+1 k k−1 ak+1 = a1 + d1 + d2 + d3 + d1 3 3 3 ¹ º ¹ º ¹ º 3m − 2 + 1 3m − 2 3m − 2 − 1 = a1 + d1 + d2 + d3 + d1 3 3 3 = a1 + (m − 1)d1 + (m − 1)d2 + (m − 1)d3 + d1 ¹ º ¹ º ¹ º 3m 2 1 = a1 + d1 + m − 1 + d2 + m − 1 + d3 3 3 3 ¹ º ¹ º ¹ º (k + 1) + 1 k+1 (k + 1) − 1 = a1 + d1 + d2 + d3 3 3 3 ∴ P(k + 1) holds for k = 3m − 2. (ii.) If k = 3m − 1 , where m ∈ N , then ak+1 = ak + d2 ¹ º ¹ º ¹ º k+1 k k−1 ak+1 = a1 + d1 + d2 + d3 + d2 3 3 3 ¹ º ¹ º ¹ º 3m − 1 + 1 3m − 1 3m − 1 − 1 = a1 + d1 + d2 + d3 + d2 3 3 3

Sequence of numbers with three alternate...

261

= a1 + md1 + (m − 1)d2 + (m − 1)d3 + d2 ¹ º ¹ º ¹ º 1 3m 2 = a1 + m + d1 + d2 + m − 1 + d3 3 3 3 º ¹ º ¹ º ¹ k+1 (k + 1) − 1 (k + 1) + 1 d1 + d2 + d3 = a1 + 3 3 3 ∴ P(k + 1) holds for k = 3m − 1. (iii.) If k = 3m , where m ∈ N , then ak+1 = ak + d3 ¹ º ¹ º ¹ º k+1 k k−1 ak+1 = a1 + d1 + d2 + d3 + d3 3 3 3 º ¹ º ¹ º ¹ 3m + 1 3m 3m − 1 = a1 + d1 + d2 + d3 + d3 3 3 3 = a1 + md1 + md2 + (m − 1)d3 + d3 ¹ º ¹ º ¹ º 2 3m 1 = a1 + m + d1 + m + d2 + d3 3 3 3 º ¹ º ¹ º ¹ k+1 (k + 1) − 1 (k + 1) + 1 d1 + d2 + d3 = a1 + 3 3 3 ∴ P(k + 1) holds for k = 3m. Therefore, (1) holds when n = k + 1. This proves the theorem. Theorem 1.4. The formula of the general term of an can also be ¹ º µ¹ º ¹ º¶ µj k ¹ º¶ n−1 n+1 n−1 n n−1 an = a1 + − d+ − d1 + d2 (2) 3 3 3 3 3 where d = d1 + d2 + d3 . Formula (2) can be shown easily using induction on n. The proof for the theorem is ommited. Now we proceed to the sum of the first n terms of the sequence. Theorem 1.5. The sum of the of the first n terms of the sequence, denoted by Sn , is given by º µ¹ º 2 ¹ jnk ¶ 1 X n+i n+1 d3 Sn = na1 + d +2 d1 − 2 i=0 3 3 3 where d = d1 + d2 + d3

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Proof. Let d = d1 + d2 + d3 . Sn = a1 + (a1 + d1 ) + (a1 + d1 + d2 ) + (a1 + d1 + d2 + d3 ) +(a1 + 2d1 + d2 + d3 ) + (a1 + 2d1 + 2d2 + d3 ) + . . . º º ¶ µ ¹ ¹ º ¹ k+1 k k−1 + a1 + d1 + d2 + d3 3 3 3 = (a1 + (1 − 1)d) + (a1 + d1 + (1 − 1)d) + (a1 + d1 + d2 + (1 − 1)d) +(a1 + (2 − 1)d) + (a1 + d1 + (2 − 1)d) + (a1 + d1 + d2 + (2 − 1)d) ³ ³j n k ´ ´ +(a1 + (3 − 1)d) + . . . + a1 + d1 + d2 + −1 d 3 µ µ¹ º ¶ ¶ µ µ¹ º ¶ ¶ n+2 n+1 − 1 d + a1 + −1 d + a1 + d1 + 3 3 µ¹ º ¹ º j k¶ ¹ º µ¹ º ¶ n+2 n+1 n 1 n+2 n+2 + + a1 + −1 d = 3 3 3 2 3 3 ¹ º ¹ º µ¹ º ¶ jnk n+1 n+1 1 n+1 + d1 + −1 d+ (d1 + d2 ) 3 2 3 3 3 ´ 1 j n k ³j n k −1 d + 2 3 µ¹3 º µ¹ º ¶ ¹ º µ¹ º ¶¶ n+2 n+2 n+1 n+1 1 = na1 + −1 + −1 d1 2 3 3 3 3 µ ¹ º j n k j n k ³j n k ´¶ n+1 1 + 2 +2 + −1 d1 2 3 3 3 3 µ¹ º µ¹ º ¶ ¹ º µ¹ º ¶¶ n+2 n+2 n+1 n+1 1 −1 + −1 d2 + 2 3 3 3 3 µ¹ º µ¹ º ¶ ¹ º µ¹ º ¶¶ n+2 n+2 n+1 n+1 1 + −1 + −1 d3 2 3 3 3 3 ´´ ´´ 1 ³j n k ³j n k 1 ³ j n k j n k ³j n k + 2 + − 1 d2 + − 1 d3 2 3µ¹ 3 º¹ 3 º ¹ 2º ¹ 3 º3 j k ¹ º¶ 1 n+2 n−1 n+1 n+4 n n+3 = na1 + + + d1 2 3 3 3 3 3 3 µ¹ º¹ º ¹ º¹ º j k¹ º¶ 1 n+2 n−1 n+1 n−2 n n+3 + + + d2 2 3 3 3 3 3 3 µ¹ º¹ º ¹ º¹ º j k¹ º¶ 1 n+2 n−1 n+1 n−2 n n−3 + + + d3 2 3 3 3 3 3 3 µ¹ º µ¹ º ¶ ¹ º µ¹ º ¶¶ n+2 n+2 n+1 n+1 1 −1 + −1 d = na1 + 2 3 3 3 3 µ¹ º ´ jnk ¶ 1 j n k ³j n k n+1 + −1 d+2 d3 d1 − 2 3 3 3 3


Lemma 1.6. For any positive integers p, q, and n, · ¸ · ¸ p p + nq +n= q q Proof. · ¸ p p ⇒ k ≤ < k + 1 where k is an integer q q p ⇒ m ≤ + n < m + 1, m = n + k. q · ¸ · ¸ p p + nq ∴ +n= . q q

Theorem 1.7. For any integer m > 0 º j k³ n ¹ j n k´ X i n = n+1−m m m m i=mq where q =

¥n¦ . m

Proof. º n ¹ X i = m i=mq = = = = =

n−mq ¹

º i + mq m i=0 ¹ º¶ n−mq µ X i q+ m i=0 n−mq ¹ X i º n−mq X + q m i=0 i=0 ¹ º ¹ º ¹ º 1 n − mq 0 + + ... + + q(n + 1 − mq) m m m ¹ º ¹ º jnk 0 1 − q + q(n + 1 − mq) + + ... + m m m j n k³ j n k´ n+1−m m m X

Corollary 1.8. For any integer m > 0, º j kµ ¹ º¶ n ¹ X m n+m n i n+1− = m m 2 m i=0

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Proof. Let q =

jnk

m n ¹ X i=0

i m

º

º 2m−1 X¹iº i = + + ... m m i=0 i=m ¸ X º mq−1 · n ¹ X i i + + m m i=mq i=m(q−1)   º º (j+1)m−1 ¹ q−1 n ¹ X X X i i  + = m m i=jm i=mq j=0 m−1 X¹

= = = =

º n ¹ X i mj + m j=0 i=mq mq (q − 1) + q(n + 1 − mq) 2³ ´ mq m − + n + 1 − mq q 2 2 ¹ º¶ j n kµ m n+m n+1− m 2 m q−1 X

Theorem 1.9. The sum of the first n terms of the sequence can also be ¹ ºµ ¹ º¶ ¹ º¶ jnk µ n+1 3 n+4 3 n+3 Sn = na1 + n+2− d1 + d2 n+1− 3 2 3 3 2 3 ¹ ºµ ¹ º¶ n−1 3 n+2 + n− d3 3 2 3 Proof. Sn

¹ º ¹ º ¹ º ¶ n µ X i+1 i i−1 a1 + = d1 + d2 + d3 3 3 3 i=1 º º n ¹ n ¹ º n ¹ X X X i+1 i i−1 = na1 + d1 + d2 + d3 3 3 3 i=1 ¹i=1 ºµ ¹ º¶i=1 n+1 3 n+4 = na1 + n+2− d1 3 2 3 ¹ º¶ ¹ ºµ ¹ º¶ jnk µ 3 n+3 n−1 3 n+2 n+1− d2 + n− d3 + 3 2 3 3 2 3

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2

Geometric sequence of numbers with three alternate common ratios

Definition 2.1. A sequence of numbers {an } is called a sequence of numbers with three alternating common ratios if the following conditions are satisfied: a3k−1 = r1 , a3k−2 a3k (ii) for all k ∈ N , = r2 , a3k−1 a3k+1 (iii) for all k ∈ N , = r3 , a3k (i) for all k ∈ N ,

where r1 , r2 , and r3 are called the first, the second and the third common ratios of {an } respectively. Example 2.2. The number sequence 1, 1/2, 1/6, 1/24, 1/48, 1/144, 1/576, 1/1152, 1/3456, . . . is an example of the sequence where r1 = 1/2, r2 = 1/3, and r3 = 1/4. Obviously, {an } has the following form a1 , a1 r1 , a1 r1 r2 , a1 r1 r2 r3 , a1 r12 r2 r3 , a1 r12 r22 r3 , a1 r12 r22 r32 , a1 r13 r22 r32 , . . . Theorem 2.3. The formula of the general term of an is e

e

an = a1 · r1n+1 · r2en · r3n−1 where ei =

¥i¦

Proof. Let ei =

3

(3)

.

¥i¦ 3

and use induction on n to prove theorem 2.3.

Obviously, (3) holds for n = 1, 2, 3 and 4. Now suppose (3) holds when n = k, hence e

e

ak = a1 · r1k+1 · r2ek · r3k−1 We need to show that P(k + 1) also holds for any k ∈ N .

(i.) If k = 3m − 2 , where m ∈ N , then ak+1 = ak · r1

(4)

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e

e

ak = a1 · r1k+1 · r2ek · r3k−1 · r1 e e e = a1 r13m−2+1 r23m−2 r33m−2−1 · r1 = a1 r1m−1 r2m−1 r3m−1 · r1 b 3m c bm−1+ 23 c bm−1+ 13 c r3 = a1 r1 3 r2 (k+1)+1 k+1 b c b 3 c b (k+1)−1 c = a1 r1 3 r2 r3 3 ∴ P(k + 1) holds for k = 3m − 2. (ii.) If k = 3m − 1 , where m ∈ N , then ak+1 = ak · r2

e

e

ak = a1 · r1k+1 · r2ek · r3k−1 · r2 e e e = a1 r13m−1+1 r23m−1 r33m−1−1 · r2 = a1 r1m r2m−1 r3m−1 · r2 bm+ 1 c b 3m c bm−1+ 23 c = a1 r1 3 r2 3 r3 b (k+1)+1 c b k+1 c b (k+1)−1 c = a1 r1 3 r2 3 r3 3 ∴ P(k + 1) holds for k = 3m − 1. (iii.) If k = 3m , where m ∈ N , then ak+1 = ak · r3

e

e

ak = a1 · r1k+1 · r2ek · r3k−1 · r3 e e = a1 r13m+1 r2e3m r33m−1 · r3 = a1 r1m r2m r3m−1 · r3 bm+ 2 c bm+ 1 c b 3m c = a1 r1 3 r2 3 r3 3 b (k+1)+1 c b k+1 c b (k+1)−1 c = a1 r1 3 r2 3 r3 3 ∴ P(k + 1) holds for k = 3m. Therefore, (5) holds when n = k + 1 and this proves the theorem. Theorem 2.4. The formula of the general term of an can also be e

an = a1 ren−1 r1n+1 ¥ ¦ where r = r1 · r2 · r3 and ei = mi .

−en−1

e −en−1

r2n


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The proof for theorem 2.4 is ommited but it can be easily verified using mathematical induction. Theorem 2.5. The formula for the sum of the first n terms of the sequence is given by µ µ¹ µ µ ¶ ¶ º j k¶¶ n+1 1 − ren−1 n en−1 r1 Sn = a1 R + 1 + a1 r − 1−r 3 3 º¶¶ µ µj k ¹ n n−1 en−1 +a1 r − (r1 + r1 r2 ) 3 3 ¥ ¦ where R = r1 + r1 r2 + r1 r2 r3 , r = r1 r2 r3 and en−1 = n−1 3 ¥ ¦ Proof. Let p = en−1 = n−1 , R = r1 + r1 r2 + r1 r2 r3 and r = r1 r2 r3 . 3 Sn = a1 + a1 r1 + a1 r1 r2 + a1 r1 r2 r3 + a1 r12 r2 r3 + a1 r12 r22 r3 + a1 r12 r22 r32 e e e +a1 r13 r22 r32 + a1 r13 r23 r32 + a1 r13 r23 r33 + . . . + a1 r1n−1 r2n−2 r3n−3 e e e e +a1 r1en r2n−1 r3n−2 + a1 r1n+1 r2en r3n−1 e e = a1 + a1 R + a1 rR + a1 r2 R + . . . + a1 rp−1 R + a1 r1en r2n−1 r3n−2 e e +a1 r1n+1 r2en r3n−1 µ¹ º j k¶ ¡ ¢ n+1 n 2 p−1 p = a1 + a1 R 1 + r + r + . . . + r + a1 r1 r − 3 3 µj k ¹ º¶ n n−1 +a1 rp (r1 + r1 r2 ) − 3 3 ¶ µ¹ º j k¶ µ p n+1 1−r n p + a1 r1 r − = a1 + a1 R 1−r 3 3 µj k ¹ º¶ n−1 n − +a1 rp (r1 + r1 r2 ) 3 3

References [1] Zhang Xiong and Zhang Yilin, ”Sequence of numbers with alternate common differences”, Scientia Magna, High American Press Vol.3, No.1, (2007), pp.93-97.