Mathematical Investigations IV
Sample Test on Sequences and Series
Name:
Key
No calculators necessary today, so please put them away. In some cases, you will be asked to
give a final answer as a single number. In other cases, you will be asked to set up the problem,
supplying numbers, but not doing the arithmetic to obtain a single number as your answer.
Please read the instructions carefully. Answers only may not receive full credit.
These formulas may be useful:
n
n(n  1)(2n  1)
j2 

6
j 1
(1)
n

j3 
j 1
n2 (n  1) 2
4
In an arithmetic sequence, find a5 if a1 = 23.4 and a41 =263.4.
263.4  23.4
6
41  1
So, a5  23.4  6(5  1)  47.4
d
(2)
Find the missing terms in the following sequences.
(a) the geometric sequence
an  a1  r n 1 
1
,
3
,
,
125
24
125 1 3
125
5
 r  r3 
 r 3  . So the sequence becomes:
24 3
8
2
1 5 25 125
, , ,
3 6 12 24
(b) the harmonic sequence
2
,
15
,
,
1
3
15
15
6
, ___, ____,3 or
, ___, ____, .
2
2
2
15 12 9 6
3
, , , where d   .
One can see that the missing terms are
2 2 2 2
2
2 2 2 2
2 1 2 1
, , , or
, , ,
So the harmonic sequence is:
15 12 9 6
15 6 9 3
We look at the associated arithmetic sequence:
p.1
Fall 08
(3)
Express the following using sigma notation. (Do NOT evaluate.)
2 6 18 54 162 486 1458 4374 8 ( 1) k  (2  3k 1)
    




5 9 13 17 21 25
29
33
1  4k
k 1
(4)
Find the value of
200
 j.
(Evaluate and simplify to a single number. Show work.)
j  21
200

j  21
200
20
j 1
j 1
j  j  j 
(200)(201) (20)(21) 20

  2010  21  10  1989  19890
2
2
2
  2k
50
(5)
Find the value of
2
k 1
 7 .
(Express with numbers, without , so that it could be
evaluated on a simple calculator. Do not do final arithmetic.)
50
50
 2k 2  7   2 k 2   7  2  50  516 101  50  7

k 1
k 1
k 1
50
12
(6)
Find the value of
3
4
k 1
k
.
(Express with numbers, without , so that it could be
evaluated on a simple calculator. Do not do final arithmetic.)
3  1
1   
12
3 4   4 


k
1
k 1 4
1
4
12



p.2
Fall 08
(8)
Find the infinite sum, if possible.
This is an infinite series with r  
3 1 1 1
   
2 2 6 18
1
3
and first term a 
3
2
3
9
2


 1 8
1  
 3
1
1
1 
1
1
 1
2  8  1 1  3

 n  n 3    n   n 3 
1
1
2  n 1 2
4 4
n 1  2
n 1 2
1
1
2
2

(9)
3 1 1 1
   
2 2 6 18


p.3
Fall 08
(10)
Write 0.1444… as an infinite series and then as a fraction.
1
4
4



10 100 1000
1  4
4
 


10  100 1000
.144444 



 4 
1  100  1
4 13
 
 


10  1  1  10 90 90
 10 

k
 3A 
(11) Find all values of A for which the infinite series  5  
 has a finite sum.
 2 
k 1
3A
To converge, we need 1  r  1 . In this case r 
. So,
2
3A
2
2
1 
1
 A
2
3
3
(12)
Suppose you have 800 grams of sand in a red bucket. You take half of the sand from the
red bucket, and transfer it to the blue bucket.
Now, you take half of the sand remaining in the red bucket and transfer it to the blue
bucket. You continue this pattern, always transferring half of the remaining sand from the
red bucket into the blue bucket.
Let bn represent the weight in grams of sand transferred to the blue bucket on the
nth transfer. For example, on the first transfer, you moved 400 grams of sand into the blue
bucket, so b1 = 400.
Write the terms in the sequence bn n 1 . 400,200,100,50,25
5
 400 if n  1

Write a recursive formula for the sequence bn n 1 . bn   1
 2  bn 1 if n  1

n
Consider the series Sn   bk . In one or two sentences, give an interpretation of S n in
k 1
n 1
1
 400    . This is the amount of
2
sand (in grams) that is in the blue bucket after n transfers. For example, after one transfer,
there are 400 grams of sand in the blue bucket. After 4 transfers, there are 750 grams of
sand in the bule bucket (i.e. 750 = 400 + 200 + 100 + 50).
terms of sand and buckets. S n  400  200  100 
p.4
Fall 08