Question

sum_(k=1)^6-3(2)^k-1

Answer 1

\[\sum_{k=1}^6{3(-2)^k-1}\] The -1 is up with the k. Find the indicated sum of the geometric series.

Answer 2

\[\sum_{k=1}^6{3(-2)^{k-1}}\]

Answer 3

Find the indicated sum of the geometric series.

Answer 4

with so few terms i would just add them

Answer 5

How so? And I messed up. It is -3(2). Sorry

Answer 6

\[\sum_{k=1}^6{3(-2)^{k-1}}=3\sum_{k=1}^6{(-2)^{k-1}}\]
\[=3(1-2+4-8+16-32)\]

Answer 7

\[\sum_{k=1}^6-3(2)^{k-1}\]

Answer 8

like that?

Answer 9

yes

Answer 10

\[\sum_{k=1}^6-3(2)^{k-1}=-3\sum_{k=1}^6(2)^{k-1}\]

Answer 11

\[-3\sum_{k=1}^6(2)^{k-1}=-3(1+2+4+8+16+32)=-3(63)\] if my arithmetic is correct

Answer 12

Yes! Thank you so much satellite73!!!! :)

Answer 13

yw