Identify the common ratio of the geometric series.

3a) Find the range of value of x for which the series _r=1^ ((3)/(x+2))^r converges. Step 1: Identify the common ratio of the geometric series. A geometric series _r=1^ ar^r-1 converges if and only if the common ratio r satisfies |r| < 1. For the given series _r=1^ ((3)/(x+2))^r, the common ratio is r = (3)/(x+2). Step 2: Set up the inequality for convergence. |(3)/(x+2)| < 1 Step 3: Solve the inequality. This inequality can be rewritten as 3 < |x+2|. This implies two separate inequalities: x+2 > 3 or x+2 < -3 Step 4: Solve each inequality for x. For the first inequality: x+2 > 3 x > 3-2 x > 1 For the second inequality: x+2 < -3 x < -3-2 x < -5 The range of values for x for which the series converges is x < -5 or x > 1. The range of values for x is x < -5 or x > 1. 3b) The sum of the first n terms of a series is S_n = 5(2^n - 1). Find the general term (a_n) and show that the sequence is a G.P. Step 1: Find the general term a_n. The general term a_n is given by a_n = S_n - S_n-1 for n > 1, and a_1 = S_1. First, find a_1: S_1 = 5(2^1 - 1) = 5(2-1) = 5(1) = 5 So, a_1 = 5. Next, find S_n-1: S_n-1 = 5(2^n-1 - 1) Now, find a_n for n > 1: a_n = S_n - S_n-1 a_n = 5(2^n - 1) - 5(2^n-1 - 1) a_n = 5 [ (2^n - 1) - (2^n-1 - 1) ] a_n = 5 [ 2^n - 1 - 2^n-1 + 1 ] a_n = 5 [ 2^n - 2^n-1 ] a_n = 5 [ 2 · 2^n-1 - 2^n-1 ] a_n = 5 [ (2-1) · 2^n-1 ] a_n = 5 · 2^n-1 This formula also holds for n=1 (a_1 = 5 · 2^0 = 5). The general term is a_n = 5 · 2^n-1. Step 2: Show that the sequence is a Geometric Progression (G.P.). A sequence is a G.P. if the ratio of consecutive terms is constant. This constant ratio is the common ratio, r. We need to show that (a_n)/(a_n-1) is a constant for all n > 1. Using the general term a_n = 5 · 2^n-1: a_n-1 = 5 · 2^(n-1)-1 = 5 · 2^n-2 Now, calculate the ratio: (a_n)/(a_n-1) = 5 · 2^n-15 · 2^n-2 = 2^n-12^n-2 = 2^(n-1) - (n-2) = 2^n-1-n+2 = 2^1 = 2 Since the ratio of consecutive terms is a constant value of 2, the sequence is a Geometric Progression.