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Discrete Mathematics Math 6A

Discrete Mathematics Math 6A. Homework 8 Solution. 3.2-2 128 b) 7 c) 2 d) -256 3.2-3 a0=2, a1=3, a2=5, a3=9 a0=1, a1=4, a2=27, a3=256 a0=0, a1=0, a2=1, a3=1 a0=0, a1=1, a2=2, a3=3 3.2-13 2+3+4+5+6=20 1-2+4=8+16=11 3+3+...+3 = 10*3 = 30

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Discrete Mathematics Math 6A

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  1. Discrete MathematicsMath 6A Homework 8 Solution

  2. 3.2-2 • 128 b) 7 c) 2 d) -256 • 3.2-3 • a0=2, a1=3, a2=5, a3=9 • a0=1, a1=4, a2=27, a3=256 • a0=0, a1=0, a2=1, a3=1 • a0=0, a1=1, a2=2, a3=3 • 3.2-13 • 2+3+4+5+6=20 • 1-2+4=8+16=11 • 3+3+...+3 = 10*3 = 30 • (2-1)+(4-2)+(8-4)+...+(512-256) = 511

  3. 3.2-16 • a) 2+0+2+0+2+0+2+0+2=10 • {(3^9-1)/(3-2)} – {(2^9-1)/(2-1)} = 9330 • (2*(3^9)-2)/(3-1) – (3*(2^9)-3)/(2-1) = 21215 • 2^9 – 2^0 = 511 • 3.2-23 • 200*201/2 – 99*100/2 = 15150 • 3.2-24 200^2*201^2/4 – 98^2*99^2/4 = 380,477,799

  4. 3.2-31 • countable: 1 > -1, 2> -2 , 3> -3... • countable: 1>0, 2>2, 3> -2, 4>4, ... • not countable • countable • 3.2-33 Yes, not countable • 3.3-3 p(n)=3+3*5+...+ 3*5^n = (5^n+1 -1)/4 • p(0) = 3 • p(k+1) = 3+3*5+...+ 3*5^k + 3*5^k+1 = (5^k+2 -1)/4 • p(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)/4 • p(0) = 2 • p(k+1) = 2-2*7 + 2*7^2 - ...+2(-7)^k + 2(-7)^k+1 = (1-(-7)^k+2)/4 • p(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)

  5. 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

  6. 3.3-45 (xy)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)

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