Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.22

22. Conjecture a formula for the nth term of {a_n}  if the first ten terms of this sequence are as follows.

a) 2, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366
b) 1, 1, 0, 1, 1, 0, 1, 1, 0, 1
c) 1, 2, 3, 5, 7, 10, 13, 17, 21, 26
d) 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365

Solution:

(a) 2, 6, 18, 54,162,486,1458,4374,13122,39366

We can see that this sequence is neither in arithmetic progression nor in geometric Progression. We use the method of inspection to find the

nth term of this sequence.

2=2(3^0)
6=2(3^1)
18=2(3^2)
54=2(3^3)
162=2(3^4)
486=2(3^5)
1458=2(3^6)
4374=2(3^7)
13122=2(3^8)
39366=2(3^9)

Therefore we have for n=1,2,...,10

a_n=2(3^{n-1})

(b) 1,1,0,1,1,0,1,1,0,1

This sequence contains 0’s and 1’s only. We notice that 0 occurs in third multiple term of the sequence.
Therefore, we have for n=1,2,...,10

a_n = \left\{\begin{matrix}0; n=3m&for& m=1,2,3 & \\1; otherwise & \end{matrix}\right.

(c) 1, 2, 3,5,7,10,13,17,21,26

We can see that this sequence is neither in arithmetic progression nor in geometric Progression. We split the sequence into odd and even terms and use the method of inspection to find the nth term of this sequence. Now, for 1,3,7,13,21 we have

1=1^2-0
3=2^2-1
7=3^2-2
13=4^2-3
21=5^2-4

So the odd termed sequence is of the form

a_{2n-1} = n^2-(n-1)
Where n=1,2,3,4,5
Similarly, for 2,5,10,17,26 we have

2=1^2+1
5=2^2+1
10=3^2+1
17=4^2+1
26=5^2+1

So the even termed sequence is of the form

a_{2n}=n^2+1
where n=1,2,3,4,5
Therefore, we have for n=1,2,…,10

a_n = \left\{\begin{matrix} a_{2m-1} = m^2-(m-1),m=1,2,3,4,5 & \\ a_{2m} = m^2 + 1,m=1,2,3,4,5 & \end{matrix}\right.

(d) 3, 5,11,21,43,85,171,341,683,1365

We can see that this sequence is neither in arithmetic progression nor in geometric Progression. We use the method of inspection to find the nth term of this sequence.

5=2.3-(-1)^2
11=2.5-(-1)^3
21=2.11-(-1)^4
43=2.21-(-1)^5
85=2.43-(-1)^6
171=2.85-(-1)^7
341=2.171-(-1)^8
683=2.341-(-1)^9
1365=2.683-(-1)^10

Therefore we have for n=1,2,….,10

a_1=3,a_n=2a_{n-1}-(-1)^n, n \geq 2

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