Question 1151137
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Answer: *[Tex \LARGE \displaystyle {\color{red}{\sum_{k=1}^{4}10(2)^{k-1}}}]


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More explanation:


If you plugged in k = 1, you'd get *[Tex \large 10(2)^{k-1} = 10(2)^{1-1} = 10] as the first term.
If you plugged in k = 2, you'd get *[Tex \large 10(2)^{k-1} = 10(2)^{2-1} = 20] as the second term.
If you plugged in k = 3, you'd get *[Tex \large 10(2)^{k-1} = 10(2)^{3-1} = 40] as the third term.
If you plugged in k = 4, you'd get *[Tex \large 10(2)^{k-1} = 10(2)^{4-1} = 80] as the fourth term.


So,  *[Tex \large \displaystyle \sum_{k=1}^{4}10(2)^{k-1} = 10 + 20 + 40 + 80]


The sequence 10, 20, 40, 80, ... is geometric with 
a = first term = 10
r = common ratio = 2
The general nth term of a geometric sequence is *[Tex \large a(r)^{n-1}]
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