Question
Prove the following proposition using induction:
P(n): 1² + 2² + 3² + 4² + ... ... ... ... . + n² = n(n+1)(2n+1) / 6
Answer :
Word Count : 310
For n = 1, the left-hand side is 1² = 1. The right-hand side gives 1(1 + 1)(2·1 + 1) / 6 = 1·2·3 / 6 = 1. Thus the proposition is true for n = 1. Assume that the proposition is true for some positive integer k, that is, 1² __________ ______ _____ _________ ____ _____ _________ ______.
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For n = 1, the left-hand side is 1² = 1. The right-hand side gives 1(1 + 1)(2·1 + 1) / 6 = 1·2·3 / 6 = 1. Thus the proposition is true for n = 1. Assume that the proposition is true for some positive integer k, that is, 1² __________ ______ _____ _________ ____ _____ _________ ______.
______ _____ ___ _____ ____ _________.
______ __________ _________ ___ ___ ______ ________ __________ ________ _________ ____.
__________ _______ __________ _________ ____ __________ ______ ______.
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_______ _________.
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Prove the following proposition using induction:
P(n): 1² + 2² + 3² + 4² + ... ... ... ... . + n² = n(n+1)(2n+1) / 6
Insertion sort
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. For the function defined by f(n) = 2n³ + 3n² + 1 and g(n) = 2n² +3, show that
(i) f(n) = Ω(g(n))
(ii) g(n) != Ω(f(n))
(iii) n³ = Ω(g(n))
(iv) f(n) != Ω(n⁴)
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