mathmath333
  • mathmath333
find the \(\large 1031^{th}\) term of the sequence.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} 1,\ 2,\ 2,\ 4,\ 4,\ 4,\ 4,\ 8,\ 8,\ 8,\ 8,\ 8,\ 8,\ 8,\ 8,\cdots \ \infty +\hspace{.33em}\\~\\ \end{align}}\)
mathmate
  • mathmate
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?
mathmath333
  • mathmath333
no i didnt understand

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kanwal32
  • kanwal32
1024
kanwal32
  • kanwal32
or 512
mathmate
  • mathmate
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?
mathmath333
  • mathmath333
is it 1024
mathmate
  • mathmate
Yes, the next one is 1024! In fact, the 1024th term starts at 1024, etc.

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