Exam 2 CMSC 203 Fall 2004
1. Circle T if the corresponding statement is True or F if it is False.
TFEvery integer greater than 1 is divisible by a prime.
TFFor all integers a, b, and c, if b divides a and c divides a, then GCD(b,c) = 1.
TFEvery iterative algorithm can be expressed as a recursive algorithm.
TF1 + 2 + 3 + ... + 1,000,000 = 500,000(1,000,000).
TFFor integers a and b, if a and b are relatively prime, then GCD(a,b) = 1.
TFThe Fibonacci Sequence is 1, 1, 2, 3, 5, 8, 13, 21, ... .
TF(31000 โ 3999) + (3999 โ 3998) + ... + (33 โ 32) + (32 โ 3) + ( 3 โ 1) is a Telescoping Series.
TFFor any non-zero integers a and b, ab = GCD(a,b)LCM(a,b).
2. (a) Given the recursion defined as f(n) = f(n
โ
1) + 2f(n
โ
2) + 3f(n
โ
3), find the next 4 terms
when f(0) = 0, f(1) = 1 and f(2) =
โ
1
(b) Given the sequences, {an} and {bn}, defined for n > 0, by: an = 2n and bn = n2,
find c0, c1, c2, and c3 when cn = an + bn.
3. List out an algorithm to apply the Quotient-Remainder Theorem to integers n and d.
4. The following algorithms evaluate the polynomial anxn + an โ 1xn โ 1 +...+ a1x + a0 at x = c. Step
through each to evaluate 2x3 + x2 + 2x + 1 at x = 3.
procedure poly1 (c, a0, a1, ..., an : real numbers)
power := 1
y := a0
for i = 1 to n
power := power * c
y := y + ai * power
next i
output y
procedure poly2 (c, a0, a1, ..., an : real numbers)
y := an
for i = 1 to n
y := y*c + an โ i
next i
output y
i 01 23
poly1 power = power = power = power =
ai = ai = ai = ai =
y = y = y = y =
poly2 an โ i = an โ i = an โ i = an โ i =
y = y = y = y =
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