Answer:
sum(2^(n+1), for n=1 to 6)
Step-by-step explanation:
To answer this question, you need to know two things:
- what is an expression for the n-th term
- how many terms are there
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The series shown is a geometric series with first term 4 and common ratio 8/4 = 2. The generic form of the n-th term is ...
an = a1×r^(n-1) . . . . first term a1, common ratio r
You can use this form directly in your summation expression, or you can simplify it a bit.
an = 4×2^(n-1) = (2^2)(2^(n-1)) = 2^(n-1+2)
an = 2^(n+1)
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The value 128 is 2^7, so n+1 = 7, or n=6 for that term
Your summation expression could be ...
[tex]\displaystyle\sum_{n=1}^6{2^{n+1}}[/tex]
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Additional comment
The n-th term can also be written as 2×2^n.
