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sarah_hendrix7

  • one year ago

Please Help?

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  1. sarah_hendrix7
    • one year ago
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  2. Hoa
    • one year ago
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    do you want the whole steps or just a result? the final result is an = -7 *6^n

  3. Hoa
    • one year ago
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    a(n+1) -6 a(n) =0 characteristic equation is x -6 =0 x =6 the equation has the form an= C*6^n replace a1 = C *6^1 = -42 ---->C = -7 replace C into an form to get an= -7 *6^n Hope this help. are you in discrete math course?If so, you will understand what i mean

  4. ParthKohli
    • one year ago
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    LOL

  5. Hoa
    • one year ago
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    What's wrong? and what makes you laugh

  6. Hoa
    • one year ago
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    @ParthKohli : I am a student and want to study from others. Please tell me if I am wrong in my work, I appreciate when you correct my mistake or my flaw

  7. ParthKohli
    • one year ago
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    You have left a lot of space. :-)

  8. Hoa
    • one year ago
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    I don't know why the net post it up. I am not good at computer

  9. ParthKohli
    • one year ago
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    I think we can do this using the geometric sequence formula.

  10. Hoa
    • one year ago
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    yes, you are right. show me your work. now is your turn

  11. ParthKohli
    • one year ago
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    Your answer is correct anyway.

  12. ParthKohli
    • one year ago
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    \(a_1 = -42\) and \(r = -6\). Now use the following formula:\[a_n = a_1 \times r^{n - 1}\]

  13. Hoa
    • one year ago
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    how to get r?

  14. ParthKohli
    • one year ago
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    Strictly speaking, if you have:\[a_{n + 1} = {\rm some ~number} \times a_n \]For all \(n\), then the "some number" is the common ratio denoted by \(r\).

  15. ParthKohli
    • one year ago
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    I'd leave it to the asker to do the rest.

  16. Hoa
    • one year ago
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    I need help from my problem, may I have yours?

  17. ParthKohli
    • one year ago
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    Sure, what may I help you with?

  18. Hoa
    • one year ago
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    my problem is: find the coefficient of x^10 in the power series of (x^2 +x^4 +....) (x^3 +x^6 +x^9 +....)(x^4 +x^8 +x^12 +....)

  19. ParthKohli
    • one year ago
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    This is an interesting problem. Let's see.

  20. ParthKohli
    • one year ago
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    You can just message me here; I check this site much often... more than I check my email

  21. Hoa
    • one year ago
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    dealt. since I have to go. send me message when you've done. Nice to see.

  22. ParthKohli
    • one year ago
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    Cya!

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