goformit100
  • goformit100
Consider the sequence 2;3;5;6;7;8;10;::: of all positive integers that are not perfect squares. Determine the 2011th term of this sequence.
Meta-math
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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goformit100
  • goformit100
@oldrin.bataku
anonymous
  • anonymous
Find the number of perfect squares under 2011 -- this is the difference between 2011 and our number.
anonymous
  • anonymous
Well, we know \(\sqrt{2011}\approx45\) with rounding (since \(44^2=1936\), \(45^2=2025\)). This means we have skipped \(45\) square numbers at this point, so our 2011th term is \(2011+45=2056\)

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