
Determine whether each of the following sequences is
arithmetic or geometric. Find the initial term, the recursion and the explicit
formula.

{a_{n}} = {2, 5, 8, 11, ... }
Solution: arithmetic. a_{1} = 2. recursion: a_{n}
= a_{n1} + 3. Explicit formula: a_{n} = 2 + 3(n1).

{b_{n}} = {2, 6, 18, 54, ... }
Solution: geometric. a_{1} = 2. recursion: a_{n}
= 3·a_{n1}. Explicit formula: a_{n} = 2·3^{n1}.

{c_{n}} = {3, 1, 5, 9, ... }
Solution: arithmetic. c_{1} = 3. recursion: a_{n}
= a_{n1}  4. Explicit formula: a_{n} = 3  4(n1).

{f_{n}} = {a_{2}, a_{5}, a_{8},
a_{11}, ... }, where a_{n} is as given above.
Solution: arithmetic. f_{1} = a_{2} = 5. recursion:
f_{n} = f_{n1} + 9. (Add 3+3+3 each time!)
Explicit formula: a_{n} = 2 + 3(n1).

In each case, use the recursion to calculate the first
six terms of each sequence:

a_{1} = 3, a_{n} = a_{n1}
+ 5 for n > 2.
solution:
n 
1 
2 
3 
4 
5 
6 
a_{n} 
3 
2 
7 
12 
17 
22 

b_{1} = 6, b_{n} = b_{n1}/
2 for n > 2.
solution:
n 
1 
2 
3 
4 
5 
6 
b_{n} 
6 
3 
3/2 
3/4 
3/8 
3/16 

p_{1} = 1, p_{2} = 2; p_{n}
= p_{n1}·p_{n2} 1 for n > 3.
solution:
n 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
p_{n} 
1 
2 
1 
1 
0 
1 
1 
0 
1 
1 

Given the recursion and initial term, find the explicit
formula:

a_{1} = 3; a_{n} = a_{n1} 
2
solution:
Explicit formula: a_{n} = 2  3(n1).

g_{1} = 7; g_{n} = 2 g_{n1}
solution:
Explicit formula: a_{n} = 7·2^{n1}.

Given the following arithmetic sequence, find the sum
(i)by
adding backward and forward,
and (ii)by using the formula 1 + 2 +
3 + ... + (n1) + n = n(n+1)/2:

{2, 5, 8, ... , 92, 95, 98}
Method 1:

2 
5 
8 
. . . 
92 
95 
98 

98 
95 
92 
. . . 
8 
5 
2 
sum 
100 
100 
100 
. . . 
100 
100 
100 
So, the sum in the table is 3300 = 33·100.
(To find the number of columns, note that
2 + 3(n 1) = 98. So, 3(n1) = 96,
or n 1 = 32.)
The sum of the sequence is 3300/2 = 1650.
Method 2: s_{n} = 33· 2 + 3(1 + 2 + ... + 32)
= 66 + 3·32·33/ 2 = 66 + 1584 = 1650.

{b_{n}} = {a_{1}, a_{2}, a_{3},... ,
a_{n}} where a_{n} = a_{1} + (k1)d for k = 1,
2, ... , n.
Method 1:

a_{1} 
a_{1} + d 
a_{1} + 2d 
. . . 
a_{1} + (n2)d 
a_{1} + (n1)d 

a_{1} + (n1)d 
a_{1} + (n2)d 
a_{1} + (n3)d 
. . . 
a_{1} + d 
a_{1} 
sum 
2a_{1} + (n1)d 
2a_{1} + (n1)d 
2a_{1} + (n1)d 
. . . 
2a_{1} + (n1)d 
2a_{1} + (n1)d 
There are ncolumns, so the sum of the table
is 2na_{1} + n(n1)d.
So, the sum of the first nterms of the sequence
is half of this, or 2na_{1} + n(n1)d/ 2.

Find the sum of the even integers starting with 500
and ending with 800.
Solution: Use the formula s_{n} = (a_{1}+a_{n})/
2. There are 151 terms (check this!!), so that the sum is 151(800+500)/
2 = 98150.

Find the sum of the first 20 terms of each of the following
sequences:

{g_{n}} = {3, 6, 12, ... }
Solution:
sum = 3(1  2^{20})/
(1) = 3(2^{20}1)

{h_{n}} = {1, 5, 25, 125, ... }
Solution:
sum = 1·(1  5^{20})/ (4) = (5^{20}1)/
4

Find the sum of the first nterms of each of the
following sequences:

{g_{n}} = {1, 2/3, 4/9, 8/27, ... }
Solution:
sum = 1·(1  (2/3)^{n})/(1
 (2/3)) = (1  (2/3)^{n})/(1/3),
so that:
Solution:
sum = (1  (2/3)^{n})·(3/1)
= 3·(1  (2/3)^{n})

{h_{n}} = {1, .9, .81, .729, ... }
Solution:
sum = 1·(1  .9^{n})/ (1  .9) =
(1  .9^{n})/ .1 = 10(1  .9^{n})

Solution:As ntends to infinity, (2/3)^{n} and
9^{n} both tend to 0. So the first sum tends to 3, and the second
sum tends to 10.