∞ ∑4(0.5)ⁿ⁻¹ n=1 For n = 1 → 4(0.5)¹⁻¹ = 4(0.5)⁰ = 4(1) = 4 For n = 2 → 4(0.5)²⁻¹ = 4(0.5)¹ = 4(0.5) = 4(0.5) For n = 3 → 4(0.5)³⁻¹ = 4(0.5)² For n = 4 → 4(0.5)⁴⁻¹ = 4(0.5)³ We notice this is a geometric series with first term: a=4 & r= 0.5 Since r<1, then the formula of the sum when n→∞ is:
Sum= a(1/1-r) Sum = 4/(1-0.5) Sum = 4/(1/2) = 4 x 2
