Derive the formula for the sum of a geometric series S_n = (a(r^n - 1))/(r-1) given the first term a = 30 and n=12 complete calculations.

The image provided is partially obscured, but based on the visible text and the context of previous questions, it appears to be asking for the derivation of the formula for the sum of a geometric series. The formula shown, r = (r(r^n - 1))/((r-1)), is likely a typo and should represent the sum S_n = (a(r^n - 1))/((r-1)), where a is the first term and S_n is the sum. The "£ 30" likely refers to the first term (a) and "12 complete cal" refers to the number of terms (n=12). Step 1: Define the sum of a geometric series. Let S_n be the sum of the first n terms of a geometric series. Let a be the first term and r be the common ratio. The terms of the series are a, ar, ar^2, , ar^n-1. The sum S_n can be written as: S_n = a + ar + ar^2 + + ar^n-1 (1) Step 2: Multiply Equation (1) by the common ratio r. Multiplying every term in Equation (1) by r: rS_n = ar + ar^2 + ar^3 + + ar^n (2) Step 3: Subtract Equation (1) from Equation (2). Subtracting the first equation from the second equation: rS_n - S_n = (ar + ar^2 + ar^3 + + ar^n) - (a + ar + ar^2 + + ar^n-1) Notice that most terms on the right-hand side cancel out: S_n(r-1) = ar^n - a Step 4: Factor out a and solve for S_n. Factor S_n from the left side and a from the right side: S_n(r-1) = a(r^n - 1) Assuming r ≠ 1, divide both sides by (r-1): S_n = (a(r^n - 1))/(r-1) This formula shows that the total value (sum) of a geometric series after n terms is given by (a(r^n - 1))/(r-1). S_n = (a(r^n - 1))/(r-1) 3 done, 2 left today. You're making progress.