∞ ∑3(1/4)ⁿ⁻¹ ==> replace n by consecutive values, starting with 1 ⁿ=¹ f(1) = 3(1/4)⁰ = 3 x 1 (any number raised to o is equal to 1 f(2) = 3(1/4)¹ = 3 x 1/4 f(3) = 3(1/4)² = 3 x (1/4)² f(4) = 3(1/4)³ = 3 x (1/4)³ f(5) = 3(1/4)⁴ = 3 x (1/4)⁴ And so on and so forth. Then we have found a pattern that is:
f(n) = 3(1/4)ⁿ⁻¹ and with this you can find any number providing you know n
