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satellite73

  • 3 years ago

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

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  1. anonymous
    • 3 years ago
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    i got my answer but apparently i was wrong this from @lopez_hatesmath

  2. anonymous
    • 3 years ago
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    maybe some fresh eyes would help

  3. anonymous
    • 3 years ago
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    nvm sorry to bother you

  4. lopez_hatesmath
    • 3 years ago
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    :(

  5. anonymous
    • 3 years ago
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    i got it

  6. lopez_hatesmath
    • 3 years ago
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    swagg

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

  8. anonymous
    • 3 years ago
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    that is, the \(a_n\) was in the NUMERATOR

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

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

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

  12. anonymous
    • 3 years ago
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    how are we doing so far?

  13. lopez_hatesmath
    • 3 years ago
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    good :)

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

  15. lopez_hatesmath
    • 3 years ago
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    okayy i got it!

  16. anonymous
    • 3 years ago
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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

  17. lopez_hatesmath
    • 3 years ago
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    gotcha thanks for the help man.

  18. anonymous
    • 3 years ago
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    yw good luck with the next one, but it works the same so you should be good to go

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