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satellite73 Group Title

find the first 4 terms \[a_1=\frac{3}{2}; a_{n+1}=\frac{n^2+1}{n(a_n)}\]

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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    i got my answer but apparently i was wrong this from @lopez_hatesmath

    • 2 years ago
  2. satellite73 Group Title
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    maybe some fresh eyes would help

    • 2 years ago
  3. satellite73 Group Title
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    nvm sorry to bother you

    • 2 years ago
  4. lopez_hatesmath Group Title
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    :(

    • 2 years ago
  5. satellite73 Group Title
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    i got it

    • 2 years ago
  6. lopez_hatesmath Group Title
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    swagg

    • 2 years ago
  7. satellite73 Group Title
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    it was this \[a_1=\frac{3}{2}; a_{n+1}=\frac{n^2+1}{n}\times a_n\]

    • 2 years ago
  8. satellite73 Group Title
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    that is, the \(a_n\) was in the NUMERATOR

    • 2 years ago
  9. satellite73 Group Title
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    so now it is not so bad replace \(n=1\) on the right hand side to get \(a_2\)

    • 2 years ago
  10. satellite73 Group Title
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    you get \[a_2=\frac{1^2+1}{1}\times \frac{3}{2}\] \[a_2=2\times\frac{3}{2}\] \[a_2=3\]

    • 2 years ago
  11. satellite73 Group Title
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    now replace \(n\) by 2 on the right hand side to get \[a_3=\frac{2^2+1}{2}\times 3\] \[a_3=\frac{5}{2}\times 3\] \[a_3=\frac{15}{2}\]

    • 2 years ago
  12. satellite73 Group Title
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    how are we doing so far?

    • 2 years ago
  13. lopez_hatesmath Group Title
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    good :)

    • 2 years ago
  14. satellite73 Group Title
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    one more? \[a_4=\frac{3^2+1}{3}\times \frac{15}{2}\] \[a_4=\frac{10}{3}\times \frac{15}{2}\] \[a_4=25\]

    • 2 years ago
  15. lopez_hatesmath Group Title
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    okayy i got it!

    • 2 years ago
  16. satellite73 Group Title
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    not so bad that is the idea, i thought the term was in the denominator which is why i was screwing it up

    • 2 years ago
  17. lopez_hatesmath Group Title
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    gotcha thanks for the help man.

    • 2 years ago
  18. satellite73 Group Title
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    yw good luck with the next one, but it works the same so you should be good to go

    • 2 years ago
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