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anonymous

  • 5 years ago

write the nth term of the sequence 1, 4, 9, 16

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  1. anonymous
    • 5 years ago
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    \[\sum_{k=1}^{n}2(k-1)+1\]

  2. anonymous
    • 5 years ago
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    What does that notation mean?

  3. myininaya
    • 5 years ago
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    the nth term is n^2

  4. myininaya
    • 5 years ago
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    since 1^2=1, 2^2=4, 3^2=9, 4^2=16,...,n^2

  5. anonymous
    • 5 years ago
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    so what would the nth term be?

  6. anonymous
    • 5 years ago
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    oh n^2 ok

  7. anonymous
    • 5 years ago
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    What about the sequence 3, 11, 31, 69, 131

  8. anonymous
    • 5 years ago
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    anyone there

  9. anonymous
    • 5 years ago
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    myininaya are you there?

  10. myininaya
    • 5 years ago
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    I'm trying to see the pattern

  11. anonymous
    • 5 years ago
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    The first differences would be 3, 20, 38, 62. The second differences would be 12, 18, 23 and 30. The third difference would be 6 for all of them

  12. anonymous
    • 5 years ago
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    Oh I meant 24 not 23

  13. anonymous
    • 5 years ago
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    Are you still there?

  14. anonymous
    • 5 years ago
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    That's funny. Both answers are correct. My answer was that the nth term was the sum of the first n odd natural numbers. So the first number is 1, the second is 1+3=4, the third is 1+3+5=9, the fourth is 1+3+5+7 = 16

  15. anonymous
    • 5 years ago
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    polpak did you get my second question?

  16. anonymous
    • 5 years ago
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    Yeah, looking at it, but these things are kinda lame cause there's a lot of different ways you can approach it.

  17. anonymous
    • 5 years ago
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    ok. tell me when you have an answer

  18. anonymous
    • 5 years ago
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    I don't get this stuff.

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