Question

Please HELP!!! What is the sum of the geometric series - asip

Answer 1

The sum of geometric series will be:
(6^(1-1))+(6^(2-1))+(6^(3-1))+(6^(4-1))
=6^0+6^1+6^2+6^3
=1+6+36+216
=259

Answer: 259

Answer 2

Answer:

259

Step-by-step explanation:

We have the next sum

$\sum_{i=1}^{4}6^{ i-1}$

Each term of the sum is

$\sum_{i=1}^{4}6^{ i-1} = 6^{1-1}+6^{2-1}+6^{3-1}+6^{4-1}$

$\sum_{i=1}^{4}6^{ i-1} = 6^{0}+6^{1}+6^{2}+6^{3}$

$\sum_{i=1}^{4}6^{ i-1} = 1 + 6 + 36 + 216$

Finally, the result is

$\sum_{i=1}^{4}6^{ i-1} = 259$