anonymous
  • anonymous
Please Help?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
do you want the whole steps or just a result? the final result is an = -7 *6^n
anonymous
  • anonymous
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

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ParthKohli
  • ParthKohli
LOL
anonymous
  • anonymous
What's wrong? and what makes you laugh
anonymous
  • anonymous
@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
ParthKohli
  • ParthKohli
You have left a lot of space. :-)
anonymous
  • anonymous
I don't know why the net post it up. I am not good at computer
ParthKohli
  • ParthKohli
I think we can do this using the geometric sequence formula.
anonymous
  • anonymous
yes, you are right. show me your work. now is your turn
ParthKohli
  • ParthKohli
Your answer is correct anyway.
ParthKohli
  • ParthKohli
\(a_1 = -42\) and \(r = -6\). Now use the following formula:\[a_n = a_1 \times r^{n - 1}\]
anonymous
  • anonymous
how to get r?
ParthKohli
  • ParthKohli
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 \(r\).
ParthKohli
  • ParthKohli
I'd leave it to the asker to do the rest.
anonymous
  • anonymous
I need help from my problem, may I have yours?
ParthKohli
  • ParthKohli
Sure, what may I help you with?
anonymous
  • anonymous
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 +....)
ParthKohli
  • ParthKohli
This is an interesting problem. Let's see.
ParthKohli
  • ParthKohli
You can just message me here; I check this site much often... more than I check my email
anonymous
  • anonymous
dealt. since I have to go. send me message when you've done. Nice to see.
ParthKohli
  • ParthKohli
Cya!

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