Discrete MathematicsMath 6A

Homework 8 Solution

3.2-2

a) 128 b) 7 c) 2 d) -256

3.2-3

a) a0=2, a1=3, a2=5, a3=9

b) a0=1, a1=4, a2=27, a3=256

c) a0=0, a1=0, a2=1, a3=1

d) a0=0, a1=1, a2=2, a3=3

3.2-13

a) 2+3+4+5+6=20

b) 1-2+4=8+16=11

c) 3+3+...+3 = 10*3 = 30

d) (2-1)+(4-2)+(8-4)+...+(512-256) = 511

3.2-16

a) 2+0+2+0+2+0+2+0+2=10

b) {(3^9-1)/(3-2)} – {(2^9-1)/(2-1)} = 9330

c) (2*(3^9)-2)/(3-1) – (3*(2^9)-3)/(2-1) = 21215

d) 2^9 – 2^0 = 511

3.2-23

a) 200*201/2 – 99*100/2 = 15150

3.2-24 200^2*201^2/4 – 98^2*99^2/4 = 380,477,799

3.2-31a) countable: 1 > -1, 2> -2 , 3> -3...b) countable: 1>0, 2>2, 3> -2, 4>4, ...c) not countabled) countable3.2-33 Yes, not countable

3.3-3 p(n)=3+3*5+...+ 3*5^n = (5^n+1 -1)/4p(0) = 3p(k+1) = 3+3*5+...+ 3*5^k + 3*5^k+1 = (5^k+2 -1)/4p(k+1) = 3*(5^k+1-1)/4 + 3*5^k+1 = (5^k+2 -1)/4

3.3-4 p(n)=2-2*7 + 2*7^2 - ...+2(-7)^n = (1-(-7)^n+1)/4p(0) = 2p(k+1) = 2-2*7 + 2*7^2 - ...+2(-7)^k + 2(-7)^k+1 = (1-(-7)^k+2)/4p(k+1) = (1-(-7)^k+1)/4 + 2*(-7)^k+1 = (1-(-7)^k+2)/4

3.3-5 p(n) = ½ + 4/1 + 1/8 + ... + ½^n = 2^n-1/2^n

p(1)=1/2

p(k+1)= 2^(k+1)-1/2^(k+1)

p(k+1) = 2^k-1/2^k – ½^(k+1)= 2^(k+1)-1/2^(k+1)

3.3-6 p(n) = 1/1*2 + 1/2*3 + ...*1/n*(n+1) = n/n+1

p(1)=1/2

p(k+1) = k+1/k+2p(k+1) = k/k+1 + 1/(k+1)(k+2) = k+1/k+2

3.3-13 n=5 2^5 = 32 > 5^2=25

Assume that 2^k > k^2 and want to derive the statement that 2^(k+1) > (k+1)^2

(k+1)^2 = k^2 +2k + 1 < k^2 +2k + k = k^2 +3k < k^2 +k^2. Thus (k+1)^2 = 2k^2 < 2*2^k

3.3-15 p(n)=1*2 + 2*3 + ...+n(n+1) = n(n+1)(n+2)/3

p(1) = 2=2

p(k+1) = {1*2 + 2*3 + ...+k(k+1)}+(k+1)(k+2) = (k+1)(k+2)(k+3)/3

p(k+1) = k(k+1)(k+2)/3 + (k+1)(k+2) = (k+1)(k+2)(k+3)/3

3.3-45 (xy)z = (x z) (y z)

p(n) = (A1 A2 ... An) B= (A1 B) (A2 B) ... (An B)

p(1) = (A1 B)

p(n+1) = (A1 A2 ... An An+1) B

= {(A1 B) (A2 B) ... (An B)} (An+1 B)

= (A1 B) (A2 B) ... (An B) (An+1 B)