Page 190 - Discrete Mathematics and Its Applications

2.4 Sequences and Summations 169

a) Set up a recurrence relation for the salary of this em- 30. Whatarethevaluesofthesesums,whereS ={1, 3, 5, 7}?
ployee n years after 2009. a) j b) j 2
b) What will the salary of this employee be in 2017? j ∈ S j ∈ S

c) Find an explicit formula for the salary of this em- c) (1/j) d) 1
ployee n years after 2009. j ∈ S j ∈ S
31. What is the value of each of these sums of terms of a
23. Find a recurrence relation for the balance B(k) owed at
the end of k months on a loan of $5000 at a rate of 7% geometric progression?
if a payment of $100 is made each month. [Hint: Ex- a) 8 3 · 2 j b) 8 2 j
press B(k) in terms of B(k − 1); the monthly interest is
j = 0 j = 1
(0.07/12)B(k − 1).]
24. a) FindarecurrencerelationforthebalanceB(k)owedat c) 8 (−3) j d) 8 2 · (−3) j
the end of k months on a loan at a rate of r if a payment j = 2 j = 0
P is made on the loan each month. [Hint: Express 32. Find the value of each of these sums.
B(k) in terms of B(k − 1) and note that the monthly 8 j 8 j j
a) (1 + (−1) ) b) (3 − 2 )
interest rate is r/12.] j = 0 j = 0
b) Determine what the monthly payment P should be so 8 8
j
j
j
that the loan is paid off after T months. c) (2 · 3 + 3 · 2 ) d) (2 j+1 − 2 )
j = 0 j = 0
25. For each of these lists of integers, provide a simple for-
mula or rule that generates the terms of an integer se- 33. Compute each of these double sums.
3 3
quence that begins with the given list.Assuming that your a) 2 (i + j) b) 2 (2i + 3j)
formula or rule is correct, determine the next three terms i = 1 j = 1 i = 0 j = 0
of the sequence. 3 2
2
3
a) 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1,... c) i d) ij
i = 1 j = 0 i = 0 j = 1
b) 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8,...
34. Compute each of these double sums.
c) 1, 0, 2, 0, 4, 0, 8, 0, 16, 0,...
2 2
d) 3, 6, 12, 24, 48, 96, 192,... a) 3 (i − j) b) 3 (3i + 2j)
e) 15, 8, 1, −6, −13, −20, −27,... i = 1 j = 1 i = 0 j = 0
f) 3, 5, 8, 12, 17, 23, 30, 38, 47,... 3 2 2 3
2
3
c) j d) i j
g) 2, 16, 54, 128, 250, 432, 686,...
i = 1 j = 0 i = 0 j = 0
h) 2, 3, 7, 25, 121, 721, 5041, 40321,... n
35. Show that (a j − a j−1 ) = a n − a 0 , where
26. For each of these lists of integers, provide a simple for- j = 1
a 0 ,a 1 ,...,a n is a sequence of real numbers. This type
mula or rule that generates the terms of an integer se-
of sum is called telescoping.
quence that begins with the given list.Assuming that your
formula or rule is correct, determine the next three terms 36. Use the identity 1/(k(k + 1)) = 1/k − 1/(k + 1) and
n
of the sequence. Exercise 35 to compute k = 1 1/(k(k + 1)).
2
2
a) 3, 6, 11, 18, 27, 38, 51, 66, 83, 102,... 37. Sum both sides of the identity k − (k − 1) = 2k − 1
b) 7, 11, 15, 19, 23, 27, 31, 35, 39, 43,... from k = 1to k = n and use Exercise 35 to find
n
c) 1, 10, 11, 100, 101, 110, 111, 1000, 1001,1010, 1011,... a) a formula for (2k − 1) (the sum of the first n
k = 1
d) 1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5,... odd natural numbers).
n
e) 0, 2, 8, 26, 80,242,728,2186,6560,19682,... b) a formula for k = 1 k.
f) 1, 3, 15, 105, 945, 10395, 135135, 2027025, ∗ 38. Use the technique given in Exercise 35, together with the
34459425,... result of Exercise 37b, to derive the formula for n k 2
k = 1
3
g) 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1,... given in Table 2. [Hint: Take a k = k in the telescoping
h) 2, 4, 16, 256, 65536, 4294967296,... sum in Exercise 35.]
∗∗ 27. Show that if a n denotes the nth positive integer that is not 39. Find 200 k. (Use Table 2.)
√ k = 100
a perfect square, then a n = n +{ n}, where {x} denotes 200 3
40. Find k . (Use Table 2.)
the integer closest to the real number x. k = 99
√
m
∗ 28. Leta n bethenthtermofthesequence1,2,2,3,3,3,4,4,4, ∗ 41. Find a formula for k = 0 k , when m is a positive
4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6,..., constructed by including integer.
√ 1 √
the integer k exactly k times. Show that a n = 2n + . ∗ m 3
2 42. Find a formula for k = 0 k , when m is a positive
29. What are the values of these sums? integer.
There is also a special notation for products. The product of
j
4
5
a) (k + 1) b) (−2)
k = 1 j = 0 a m ,a m+1 ,...,a n is represented by n a j , read as the prod-
j = m
j
8
10
c) 3 d) (2 j+1 − 2 ) uct from j = m to j = n of a j .
i = 1 j = 0