Stitz-Zeager_College_Algebra_e-book

19 we will discuss how we arrived at this

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Unformatted text preview: ow we can use P (k ) in this case to prove P (k + 1), we note that the sum in P (k + 1) is the sum of the first k + 1 terms of the sequence ak = a + (k − 1)d for k ≥ 1 while the sum in P (k ) is the sum of the first k terms. We compare both side of the equation in P (k + 1). k+1 (a + (j − 1)d) ? k+1 (2a + (k + 1 − 1)d) 2 ? k+1 (2a + kd) 2 = j =1 summing the first k + 1 terms k (a + (j − 1)d) + (a + (k + 1 − 1)d) = j =1 summing the first k terms adding the (k + 1)st term k ? (2a + (k − 1)d) +(a + kd) = 2 (k + 1)(2a + kd) 2 Using P (k) k (2a + (k − 1)d) + 2(a + kd) 2 = 2ka + k 2 d + 2a + kd 2 2ka + 2a + k 2 d + kd 2 = 2ka + 2a + k 2 d + kd 2 ? Since all of our steps on both sides of the string of equations are reversible, we conclude that the two sides of the equation are equivalent and hence, P (k + 1) is true. By the Principle of Mathematical Induction, we have that P (n) is true for all natural numbers n. 2. We let P (n) be the formula (z )n = z n . The base case P (1) is (z )1 = z 1 , which reduces to z = z which is true. We now assume P (k ) is true, that is, we assume (z )k = z k and attempt to show tha...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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