II. Series and Mathematical Induction A. Expand each summation, then simplify. (2 items x 5 points) 10 (2-5a) 2. (-1) * +2x B. Fill in the blanks to complete the proof of Mathematical Induction. (2 items x 10 points) 1. 11 + 19+ 27 + ... + (8n + 3) = n(4n + 7) for ne N Given a, = 8n + 3 and S, = n(4n + 7), the needed components are ... For an = 8n + 3: ax = 8k + 3 and ak+1 = 8k + 11 For S, = n(4n + 7): S* = k(4k + 7) and Sk+1 = (k + 1)(4k + 11). Part 1 Show S, is true for n = 1 S, = _ S= SE Part 2: Note: Each blank must be in the simplest form. Assume that Sx is true; that is 11 + 19+ 27+ ... + (8k +3) = k(4k + 7) Now show that Ske+1 is true; that is 11 + 19+ 27 + ... + (8k +3) + (8k + 11) = Using the formula for Sk and adding (8/ + 11) to both sides of the equation, we have 11 + 19+ 27 + ...+ (8k +3) + (8k + 11) =_ + Thus, the truth of Sk+1 is true, whenever Sx is true. Hence, the formula is true for all n E N. I First Porladical Exan' (n + 1)2 for n E N 4 Given an = n3 and S. = " (n+1)2 the needed components are ... For an = n3: ax = k3 and ak+1 = (k + 1)3 For S,, = "(at1)' s _ k(k+1)2 - and Skey = (k+1)?(k+2)2 Part 1 Show Sn is true for n = 1 Sin'=- 4 $1: 13 = 4 1 = Part 2: Note: Each blank must be in the simplest form. Assume that Sx is true; that is 13 + 23 + 33 + ... + 13 = - k2 (k + 1)2 4 Now show that Sk+1 is true; that is 13 + 23 + 33 + ...+ K3 + ( k + 1)3= 4 Using the formula for Sk and adding (k + 1)3 to both sides of the equation, we have 13 + 23 + 33 + ... + K3 + (k + 1)3 = . + 4 4 4 4 Thus, the truth of Sk+1 is true, whenever S, is true. Hence, the formula is true for all n E N