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mathmath333

  • one year ago

find the \(\large 1031^{th}\) term of the sequence.

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} 1,\ 2,\ 2,\ 4,\ 4,\ 4,\ 4,\ 8,\ 8,\ 8,\ 8,\ 8,\ 8,\ 8,\ 8,\cdots \ \infty +\hspace{.33em}\\~\\ \end{align}}\)

  2. mathmate
    • one year ago
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    You have already shown: 1st term=1 3rd term=2 7th term=4 15th term=8 ... can you find the 1023rd term? and hence the 1031st term?

  3. mathmath333
    • one year ago
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    no i didnt understand

  4. kanwal32
    • one year ago
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    1024

  5. kanwal32
    • one year ago
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    or 512

  6. mathmate
    • one year ago
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    In case you haven't found it yet: 1st term=\(2^1-1\) st term=1 =\(2^0\) 3rd term=\(2^2-1\) rd term=2=\(2^1\) 7th term=\(2^3-1\) th term=4=\(2^2\) 15th term=\(2^4-1\) th term=8=\(2^3\) ... can you find the 1023rd term? =\(2^{10}-1\) rd term = \(2^9\) = 512 and hence the 1031st term?

  7. mathmath333
    • one year ago
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    is it 1024

  8. mathmate
    • one year ago
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    Yes, the next one is 1024! In fact, the 1024th term starts at 1024, etc.

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