 # Lreese Homework 2.pdf - Lathaniel Reese COT3100 Professor...

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Lathaniel ReeseCOT3100Professor Bernd Owsnicki-KleweHomework 21) Prove: If n is a positive integer such that n mod 4 (the remainder of n underdivision by 4) is 2 or 3, then n is not a perfect square.With n = k^2, these four cases are considered;1.If k mod(4) = 0, then k = 4q, for some integer q. Then, n = k2= 16 q2= 4(4 q2) ,which means n mod(4) = 0.2.If k mod(4) = 1, then k = 4q + 1, for some integer q. Then, n = k2= 16 q2+ 8 q + 1=4(4 q2+ 2 q) + 1, which means n mod(4) = 1.3.If k mod(4) = 2, then k = 4q + 2, for some integer q. Then, n = k2= 16 q2+ 16 q + 4 =4(4 q2+ 4 q + 1),which means n mod(4) = 0.4.If k mod(4) = 3, then k = 4q + 3, for some integer q. Then, n = k2= 16 q2+ 24 q + 9 =4(4 q2+ 6 q + 2) + 1,which means n mod(4) = 1.If n is a perfect square, then n mod(4) is either 1 or 0. If it is not a perfect square,then n mod(4) will be the other two possible options.
2) Consider the sum 3 + 6 + 9 + ... + 3na) Guess a formula P(n) for this sum, depending on nb) Prove it by induction. [Base case: n = ? , P(k) , P(k+1) , P(k) →P(k+1)a)P(n) = 3n(n+1)/2b) Base Case: n = 13 = 33n(n+1)/2 = 3(2)/2 = 3
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