jesusfreak
sum_(k=1)^63(2)^k1



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jesusfreak
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\[\sum_{k=1}^6{3(2)^k1}\] The 1 is up with the k. Find the indicated sum of the geometric series.

anonymous
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\[\sum_{k=1}^6{3(2)^{k1}}\]

jesusfreak
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Find the indicated sum of the geometric series.

anonymous
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with so few terms i would just add them

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

anonymous
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\[\sum_{k=1}^6{3(2)^{k1}}=3\sum_{k=1}^6{(2)^{k1}}\]
\[=3(12+48+1632)\]

anonymous
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\[\sum_{k=1}^63(2)^{k1}\]

anonymous
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like that?

jesusfreak
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yes

anonymous
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\[\sum_{k=1}^63(2)^{k1}=3\sum_{k=1}^6(2)^{k1}\]

anonymous
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\[3\sum_{k=1}^6(2)^{k1}=3(1+2+4+8+16+32)=3(63)\] if my arithmetic is correct

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

anonymous
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yw