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

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

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Other answers:

with so few terms i would just add them
How so? And I messed up. It is -3(2). Sorry
\[\sum_{k=1}^6{3(-2)^{k-1}}=3\sum_{k=1}^6{(-2)^{k-1}}\] \[=3(1-2+4-8+16-32)\]
\[\sum_{k=1}^6-3(2)^{k-1}\]
like that?
yes
\[\sum_{k=1}^6-3(2)^{k-1}=-3\sum_{k=1}^6(2)^{k-1}\]
\[-3\sum_{k=1}^6(2)^{k-1}=-3(1+2+4+8+16+32)=-3(63)\] if my arithmetic is correct
Yes! Thank you so much satellite73!!!! :)
yw

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