P78:
3. What are the terms a0, a1, a2,
and a3 of the sequence {an}, where an equals
a) 2n + 1?
Answer: a0 = 2, a1
= 3, a2 = 5, a3 = 9
b) (n+1)n+1?
Answer: a0 = 1, a1
= 4, a2 = 27, a3 = 256
c) floor(n/2)?
Answer: a0 = 0, a1
= 0, a2 = 1, a3 = 1
d) floor(n/2) + ceiling(n/2)
Answer: a0 = 0, a1
= 1, a2 = 2, a3 = 3
7. What is the value of each of the following sums of terms of a
geometric progression?
a) Sum(3 * 2j), where j = 0, 1, 2, ..., 8
Answer: ( 3 * 28+1 -3 ) / ( 2 - 1 )
= 1533
b) Sum(2j), where j = 1, 2, ..., 8
Answer: ( 28+1 -1 ) / ( 2 - 1 ) - 20
= 510
c) Sum( (-3)j ), where j = 2, ..., 8
Answer: ( (-3)8+1 -1 ) / ( -3 - 1 )
- (-3)0 - (-3)1 = 4923
d) Sum( 2 * (-3)j ), where j = 0, 1, 2, ..., 8
Answer: ( 2 * (-3)8+1 -2 ) / ( -3 -
1 ) = 9842
10. Compute each of the following double sums.
a) SumOveri( SumOverj ( i - j ) ), where i = 1, 2, 3, and j = 1,
2.
Answer: (1-1) + (1-2) + (2-1) + (2-2) + (3-1) + (3-2)
= 3
b) SumOveri( SumOverj ( 3i + 2j ) ), where i = 0, 1, 2, 3, and j
= 0, 1, 2.
Answer: (3*0+2*0) + (3*0+2*1) + (3*0+2*2) + (3*1+2*0)
+ (3*1+2*1) + (3*1+2*2) + (3*2+2*0) + (3*2+2*1) + (3*2+2*2) + (3*3+2*0)
+ (3*3+2*1) + (3*3+2*2) = 78
c) SumOveri( SumOverj ( j ) ), where i = 1, 2, 3, and j = 0, 1, 2.
Answer: 3 * ( 0 + 1 + 2 ) = 9
d) SumOveri( SumOverj ( i2j3 ) ), where i
= 0, 1, 2, and j = 0, 1, 2, 3.
Answer: (02 * 03)
+ (02 * 13) + (02 *
23) + (02 * 33) +
(12 * 03) + (12 *
13) + (12 * 23) +
(12 * 33) + (22 *
03) + (22 * 13) +
(22 * 23) + (22 *
33) = 180
A-2:
2. Find each of the following quantities.
a) log2 1024
Answer: 10
b) log2 (1/4)
Answer: -2
c) log4 8
Answer: (log2 8) / (log2 4) = 3/2
4. Let a, b, and c be positive real numbers. Show that a^(logbc)
= c^(logba). Where "^" means "power".
Proof: logb c = loga c / loga
b => a^(logb c) = a^(loga c / loga b)
= (a^loga c)^(1/loga b) = c^(1/loga
b) = c^(logb a)
where, 1/loga b = 1/(logx b / logx a) = logx a / logx b = logb a
c) f(n+1) = 3 * f(n)2 - 4 * f(n-1)2f(0) = -1 f(1) = 2 f(2) = f(1)2 * f(0) = 22 * (-1) = -4 f(3) = f(2)2 * f(1) = (-4)2 * 2 = 32 f(4) = f(3)2 * f(2) = 322 * (-4) = -4096 f(5) = f(4)2 * f(3) = (-4096)2 * 32 = 536870912
d) f(n+1) = f(n-1) / f(n)f(0) = -1 f(1) = 2 f(2) = 3 * f(1)2 - 4 * f(0)2 = 3 * 22 - 4 * (-1)2 = 8 f(3) = 3 * f(2)2 - 4 * f(1)2 = 3 * 82 - 4 * 22 = 176 f(4) = 3 * f(3)2 - 4 * f(2)2 = 3 * 1762 - 4 * 82 = 92672 f(5) = 3 * f(4)2 - 4 * f(3)2 = 3 * 926722 - 4 * 1762 = 25764174848
f(0) = -1 f(1) = 2 f(2) = f(0) / f(1) = (-1) / 2 = -1/2 f(3) = f(1) / f(2) = 2 / (-1/2) = -4 f(4) = f(2) / f(3) = (-1/2) / (-4) = 1/8 f(5) = f(3) / f(4) = (-4) / (1/8) = -32
b) an = 1 + (-1)n
Answer:
an-1 = 1 + (-1)n-1
an = 1 + (-1)n = 1 + (-1)(n-1)+1
= 1 + (-1)n-1 * (-1) = 1 - (-1)n-1 =
1 - [ 1 + (-1)n-1 - 1 ]
= 1 - ( an-1 - 1 ) = 2-
an-1, for n>=1 and a1 = 0
c) an = n(n+1)
Answer:
an-1 = (n-1)n = n2 - n
an = n(n+1) = n2 + n = (n2
- n) + 2n = an-1 + 2n, for n>=1 and a1
= 2
d) an = n2
Answer:
an-1 = (n-1)2 = n2
- 2n + 1
an = n2 = ( n2
- 2n + 1 ) + 2n - 1 = an-1 + 2n - 1, for n>=1 and a1
= 1
11. Prove that f1 + f3 + ... + f2n-1
= f2n whenever n is a positive integer. ( fn is the
nth Fibonacci number.)
Proof: Prove by induction. Let P(n) be " f1
+ f3 + ... + f2n-1 = f2n
", where n = 1, 2, 3, ...