Quandaries and Queries Name: Vic Problem: Find the first 4 terms and the nth term of the infinite sequence defined recursively as follows: a(1) = 3 and a(k+1) = 2a(k) for k -> 1. Note: Quantities in brackets are subscripts -> means 'equal to or greater than'. Using the recursive formula, the first 4 terms are; a(1) = 3, a(2) = 6, a(3) = 12, a(4) = 24 The nth term a(n) = 2n-1 x 3 (equation 1) Equation 1 must be proven using mathematical induction. This is where I am having a problem. Step 1: p(1) is true since a(1) = 3 and a(n) = 2n-1 x 3 so a (1) = 20 x 3 = 3 Thus, assume P(n) is true. a(n) =2n-1 x 3 ( induction hypothesis) Step 2: Now must show that a(n+1) = 2n x 3 ? ? ? ? ? ? ? Would appreciate your assistance. Thanks Hi Vic, The recursive definition says that the first term is 3 and after that every term is twice the preceding term. You think you have an expression for a with any subscript. The expression is a (with a subscript) is (2 to the power which is one less than the subscript) times 3 You want to prove this by induction so you first show the expression is valid when the subscript is 1. That you did. For the second step you assume that the expression is valid for some particular value of the subscript and you want to show it is valid for the next value of the subscript. Thus at this point you know: I: For EVERY integer k ≥ 1, ak+1 = 2 ak (Definition) II: For some PARTICULAR integer n, an = 2n-1 3 (Induction hypothesis) So what is an+1? According to I, an+1 = 2 an According to II, an = 2n-1 3 Thus an+1 = 2 an = 2 2n-1 3 = 21 + n - 1 3 = 2n 3 Hence, by induction, the expression is valid for all subscripts which are integers greater than or equal to one. Penny Go to Math Central