# Busl250 - Mid Semester Notes

May 12th, 2015
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Hw1solFundamentals of Algorithms Fall 2010 HW 1 DUE: August 30, 20101. Prove by mathematical induction that the sum of the first N natural numbers is N* (N+ 1)/2.Base case, n = 11* (1 + 1) / 2 = 1Inductive hypothesis, n = kSum of first k natural numbers = k* (k + 1) / 2TS: sum of first k+1 natural numbers =(k+1)((k+1) + 1) / 2= (k+1)(k+2) / 2= (k2 + 3k + 2) /2Inductive step:Adding (k+1) to both sides of the IH:Sum of first k+1 natural numbers = (k(k+1) /2) + k+1= (k2 + k / 2) + k+1= k2 + k + 2k +2 / 2= (k2 + 3k +2) /2Proven!2. Prove by mathematical induction that the 2n < 3n for n >= 1.Base Case (n=1):2^1 < 3^1Inductive Hypothesis:2^k < 3^kTo Show: 2^(k+1) < 3^(k+1)Multiplying both sides of the IH by 3:2^(k+1) < 3* 2^k < 3^(k+1) QED3. Prove by mathematical induction that 1 + 3 + 5 ++ (2n-1) = n2.Base Case: 2(1)-1=1^2Inductive Hypothesis: 1+3+5+7+... (2k-1) = k^2To Show: 1+3+5+7+...+ (2k-1) + (2k+1) = (k+1)^2Add (2k+1) to both sides of the IH:1+3+5+7+... (2k-1) + (2k+1) = k^2 + (2k+1) = (k+1)^2 QED4. Prove by mathematical induction that 13 + 23 + 33 + +n3 = (N* (N+ 1)/2)2Base Case: n =1 = 1 = 1 Inductive Hypothesis: n = k + + + = To Show: n = k+1 + + + + + = = ( + + + 12k + 4)/4Add (k+1)3 to both sides of the IH:( + + + + ) + = + ( + 2k +1)(k + 1)= /4 + ( + + 3k +1)= ( + + + 12k +4)/4 = ( + + + + + 12k + 4)/4 = ( + + + 12k +

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