Question

Question #33a9c

Answer 1

`\sum_(k=0)^5 (2*(-6)^k)`

Explanation:

If you write out the prime factorization of each term, you can see a pattern:
`2 = 2`
`12 = 2*2*3`
`72 = 2*2*2*3*3`
`432 = 2*2*2*2*3*3*3`
...

You can see that each term has an extra 2 and an extra 3. So, each term is just 6 times the previous term. That makes the summation be:

`\sum_(k=0)^5 (2*(-1)^k*6^k)`

The `(-1)^k` part makes the terms alternate between adding and subtracting.
If you want to simplify the summation, you can combine the `(-1)^k` and the `6^k`:

`\sum_(k=0)^5 (2*(-6)^k)`