
Assignment
#12
Read
 Chapter 6
 Chapter 7
 Chapter 8
Problems
 Chapter 7 (pp 190195):
2,3,4,7,8,9,12,17
 Chapter 8 (pp 203205): 1d,2,3,7,9,11




Chapter 7
 #2: Based on the information in the table, f(x)
is a quadratic function because a constant
difference appears in the seconddifferences row
of the differences table.

x

1

2

3

4

5

6

f(x)

12

28

50

78

112

152

D1 >

16

22

28

34

40

D2 >

6

6

6

6

..

 #3: We expect the first constant difference to
occur in row D3 because the function is a
thirddegree polynomial. Here we see that the
first constant difference is 12.
x

1

2

3

4

5

6

f(x)

5

26

75

164

305 
510

D1 >

21 
49 
89 
141 
205 
.

D2 >

28 
40 
52 
64 
.

D3 >

12 
12 
12 
.


 #4: (See page 190 in our text for the difference
table of the general quadratic, ax^{2}+bx+c.
Row

1

2

3

4

5

6

Sum

1

5

12

22

35

51

D1 >

4 
7 
10 
13

16

.

D2 >

3 
3 
3 
3 
.

D3 >

.

.

.

.


 2a = 3 > a = 3/2; 3a + b = 4, a = 3/2
> b = 1/2; a + b + c = 1, a = 3/2, b =
1/2 > c = 0. So an explicit row sum
representation is S(n) = 3/2x^{2} 
1/2x.
 #7: See figure (1b) on page 189.
 #8: Sorry, this is a repeat of #2!
 #9: Using difference table for problem #2 and
the general quadraticpolynomial difference table
(page 190), we have:
 2a = 6 > a = 3; 3a + b = 16, a = 3 >
b = 7; a + b + c =12, a = 3, b = 7 > c = 2.
 Desired representation is 3x^{2} + 7x
+ 2.
 #12:
 The next two rows are 6,10,12,12,10,6 and
7,12,15,16,15,12,7.
 The desired function is Sum(row n) = (1/6)n^{3}+(1/2)n^{2}+(1/3)n
= (1/6)(n^{3}+3n^{2}+2n) =
(1/6)(n)(n+1)(n+2).
 S(100) = 171,700
 #17:
 Row 6: 256, 289, 324, 361, 400, 441; Row 7:
484, 529, 576, 625, 676, 729, 784
 In the rowsums difference table, a constant
difference of 30 occurs in row D5, indicating
the polynomial that explicitly represents the
row sums is a quintic polynomial (5thdegree
polynomial).
Chapter 8
 #1d: 2  x  x^{3} + 2x^{4}
 x^{5} + x^{6}  x^{7}
+ ...
 #2:

(a) AB
AB = 1 + 3x + 3x^{2} + 2x^{3}
+ 4x^{4} + 5x^{5} +
x^{6} + x^{7}

(b) AC
AC = 1 + x  x^{3} + x^{5}
+ x^{6}

(c) AD
AD = 1 + 2x + 3x^{2} + 3x^{3}
+ 3x^{4} + ...

(d) DE
DE = 1 + x + x^{2} + 2x^{3}
+ 2x^{4} + 2x^{5} +
3x^{6} + 3x^{7} +3x^{8}
+ ...

 #3: (1 + x + x^{2} + x^{3})(1
+ x + x^{2} + x^{3} + x^{4}
+ x^{5}) = 1 + 2x + 3x^{2} +
4x^{3} + 4x^{4} + 4x^{5}
+ 3x^{6} + 2x^{7} + x^{8}.
 #7: (1 + x + x^{2} + x^{3})(1
+ x^{2} + x^{4})(1 + x + x^{2}
+ x^{3} + ...) = 1 + 2x + 4x^{2}
+ 6x^{3} + 8x^{4} + 10x^{5}
+ 11x^{6} + 12x^{7} + 12x^{8}+
...
 The
coefficient of the 7thpowered term is 12,
so there are 12 ways to buy 7 pastries
 #9: (1 + x^{5} + x^{10} +
... + x^{100})(1 + x^{10} + x^{20}
+ ... + x^{100})(1 + x^{25} +
x^{50} + ... + x^{100})(1 + x^{50}
+ x^{100})
 After
expanding this product and collecting like
terms, the coefficient of the 100degree
term will be the number of ways to make
the change.
 #11:
 (a) 0 <= k <= 9 (k is between 0 and
9, inclusive)
 (b) 60 sets as described here. You can
determine this value by summing the
coefficients in the expansion of the
generating function shown below.
 (c) (1 + x + x^{2} + x^{3})(1
+ x + x^{2} + x^{3} + x^{4})(1
+ x + x^{2})
 (d) Revise the generating function to
account for the restriction: (x^{2}
+ x^{3})(x + x^{2} + x^{3}
+ x^{4})(1 + x + x^{2}).
Once expanded, look at the coefficient of
the 6thpowered term in the expansion. This
is the desired response is 6.

