Question

Part 1: Determine the geometric partial sum the sum of 3 times 2 to the I minus 1 power from 1 to 7.

Part 2: In complete sentences, explain the necessary steps required to find the partial sum. Include an explanation of whether this partial sum is divergent or convergent in nature.

Answer 1

\[\sum_{i=1}^{7}3(2)^{i-1}\]

Answer 2

The 3 can be taken out of the sigma.
Sum = 3(2^0 + 2^1 + 2^2 +.... + 2^6)
use the formula for the geometric sum.
Or in this case, since there are only 7 terms you can write each one out and add them up.

Answer 3

Sum = 3(1 + 2 + 4 + 8 + 16 + 32 + 64) = 3(127) = 381

Answer 4

I didn't mean to close this!!!! Please don't leave @ranga

Answer 5

Or, using the geometric sum formula
geometric sum = a(1 - r^n) / (1 - r)
a = 1
r = 2
n = 7
1*(1 - 2^7) / (1 - 2) = 127.
Multiply by 3 that was taken out of the sigma: 3 * 127 = 381.

Answer 6

so the sum is 381?

Answer 7

yes.

Answer 8

Thank you sooooooooooo much!!!!!!!!

Answer 9

you are welcome.