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GOODMAN

  • 3 years ago

Find the first 6 terms of each sequence: I will write down in equation editor.

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  1. imranmeah91
    • 3 years ago
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    first time I've seen you here

  2. GOODMAN
    • 3 years ago
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    \[a _{1}=1, a _{2}=2, a _{n+2}=4a _{n+1}-3a _{n}\]

  3. GOODMAN
    • 3 years ago
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    Sorry, i take forever on this equation thingie.

  4. imranmeah91
    • 3 years ago
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    ok, let's find \[a_3\] together

  5. GOODMAN
    • 3 years ago
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    Okie dokie!

  6. karatechopper
    • 3 years ago
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    its like lil emu is in dsiguise helpin ya!;D lol i will leave ya alone now to get bak to hw

  7. GOODMAN
    • 3 years ago
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    Okay. Got it.

  8. imranmeah91
    • 3 years ago
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    \[a_3=a_{1+2}\] which mean n=1 using equation \[a_{2+1}=4a_{1+1}-3a_{1}\]

  9. GOODMAN
    • 3 years ago
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    wait..so n does not equal 2?

  10. imranmeah91
    • 3 years ago
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    no, I meant n=1

  11. GOODMAN
    • 3 years ago
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    Okie.

  12. imranmeah91
    • 3 years ago
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    \[a_{2+1}=4a_{1+1}−3a_1\] \[=a_{3}=4a_{2}−3a_1\] you know what a1 and a2 are

  13. GOODMAN
    • 3 years ago
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    Oh coolio!!! We can plug those numbers in.

  14. imranmeah91
    • 3 years ago
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    a_3=4(2)-3(1)=5

  15. imranmeah91
    • 3 years ago
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    let see you find a_4

  16. GOODMAN
    • 3 years ago
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    Okay, i got 5 for a_3

  17. GOODMAN
    • 3 years ago
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    Would i have to make n equal to 2?

  18. imranmeah91
    • 3 years ago
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    yes

  19. GOODMAN
    • 3 years ago
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    I got 14 for a_4

  20. GOODMAN
    • 3 years ago
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    Is that correct?

  21. imranmeah91
    • 3 years ago
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    yes

  22. GOODMAN
    • 3 years ago
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    YAAAYYY!!! now its soo simple for me!!

  23. GOODMAN
    • 3 years ago
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    I got this, thank you thank you!!!!

  24. karatechopper
    • 3 years ago
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    done yet ??????????????

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spraguer (Moderator)
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