A geometric series has terms 3 times 2^k for k running from 1 to 6. What is its sum?

1. Substitute k = 1 to find the first term: 3 times 2 = 6. 2. The base of the exponential is 2, so the common ratio is 2. 3. The upper limit is 6, so n = 6. 4. Substitute into S_n = a_1(1 - r^n)/(1 - r) to get S_6 = 6(1 - 2^6)/(1 - 2). 5. Since 2^6 = 64, the numerator is 6(1 - 64) = 6(-63) = -378. 6. Dividing by 1 - 2 = -1 gives 378. 7. Taking the first term as 3 rather than 6 would halve the answer to 189, and 384 is 3 times 2^7, the term just past the end of the series. _Source: OpenStax Algebra and Trigonometry (CC BY 4.0), Ch 13 "Sequences, Probability and Counting Theory", section 13.4 Series and Their Notations_