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7.
P(n) : 6 divides

n3 n
Domain : {n n ≥ 0}
Basis:
P(2) :

=
23 2 6
(True)
Proof:
P(n + 1) :
( + )  ( + )
n 1 3
n 1
(Inductive Hypothesis)
=
+
+
+ 
+
n 1n2
2n 1 n 1
(Distribute)
=
+
+
+  
n3
3n2
3n 1 n 1
(Distribute)
=

+
+
n3 n
3n2
3n
(Arrange Terms)
=

+
( + )
n3 n
3n n 1
(Factor Out 3n)
Because P(n) is divisible by 6, the first term is as well.
Also, since
3n(n+1) is divisible by 3 and either (n) or (n+1) must be an even
number,
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Unformatted text preview: the second term is divisible by 6. 8. Amounts : 20, 25, 30, 35, … P(n) : . j. k(j ≥ 0 ∧ k ≥ 0 ∧ 5n = 10j + 25k) Domain : {n n ≥ 4} Basis: P(7) : True for j = k = 1 Proof: P(n + 1) : . j’. k’(j’ ≥ 0 ∧ k’ ≥ 0 ∧ 5(n + 1) = 10j’ + 25k’) Proof By Cases: Case 1 : k ≠ 0. Then j’ = j + 3 and k’ = k – 1 Case 2 : k = 0. Then j’ = j – 2 and k’ = k + 1...
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This note was uploaded on 04/28/2009 for the course CSC 226 taught by Professor Watkins during the Spring '08 term at N.C. State.
 Spring '08
 WATKINS

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