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anonymous

  • 5 years ago

Sequence Ak= 2a (k-1) - 3 if a1 =4 whats next 3 terms

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  1. anonymous
    • 5 years ago
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    K-1 is supposed to be low like a exponent but at bottom idk what its called?

  2. anonymous
    • 5 years ago
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    index A2=2*4-3=5 A3=2*5-3=7 A4=2*7-3=11 A5=2*11-3=18 and so on

  3. anonymous
    • 5 years ago
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    whats the star mean?

  4. anonymous
    • 5 years ago
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    multiply *

  5. anonymous
    • 5 years ago
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    nevermind

  6. anonymous
    • 5 years ago
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    thank you

  7. amistre64
    • 5 years ago
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    An = 2(An-1) -3 An = 2^2 (An-2) - 2(3) - 3 An = 2^3 (An-3) - 2^2(3) - 2(3) - 3 An = 2^n(A0) - 3[x] x = 1 + 2 + 2^2 + 2^3 + ... + 2^n-1 2x = 2 + 2^2 + 2^3 + ... + 2^n-1 + 2^n 2x - x = 2^n - 1 x = (2^n -1)/2 An = 2^n(A0) - 3[(2^n -1)/2] perhaps?

  8. amistre64
    • 5 years ago
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    might need to adjust for A1; tho, so lets check.... well, it dont match but i was close with it

  9. anonymous
    • 5 years ago
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    is there a way I can write it with out typing so i can show you exactly how it looks

  10. amistre64
    • 5 years ago
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    A1 = 4 A2= 2(4) - 3 = 5 A3= 2(5) - 3 = 7 A4= 2(7) - 3 = 11 is correct tho

  11. amistre64
    • 5 years ago
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    write it without typing it.... prolly not

  12. anonymous
    • 5 years ago
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    okay hold on ill use paint and up load it

  13. anonymous
    • 5 years ago
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    or like this maybe you'll understand A = 2A -3 k (k-1)

  14. anonymous
    • 5 years ago
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    but the k supposed to be up closer next to the a but set a little lower same for (k-1) Like when you do a LOG

  15. anonymous
    • 5 years ago
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    do you understand now?

  16. anonymous
    • 5 years ago
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    \[a _{k}\]

  17. anonymous
    • 5 years ago
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    yesss :)

  18. anonymous
    • 5 years ago
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    the solution is what I wrote, I cannot understand what amistre wrote :)

  19. anonymous
    • 5 years ago
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    thank you do you have time for another

  20. anonymous
    • 5 years ago
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    sure

  21. anonymous
    • 5 years ago
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    Use mathematical induction 7+ 14+ 21+ 21 +28 +.....7N= 7N(N+1)/ 2

  22. anonymous
    • 5 years ago
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    ok, do you now what induction is in maths?

  23. anonymous
    • 5 years ago
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    7n(N+1) is all divided by 2 and not really

  24. anonymous
    • 5 years ago
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    I know how to prove P(1) true but (k+1) is where im lost

  25. anonymous
    • 5 years ago
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    induction is a technique to prove things has 3 steps 1. what you wrote: prove that it is true for N=1 2. Assume that it is true for any N 3. Prove that it is true for N+1

  26. anonymous
    • 5 years ago
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    1. N=1 7=7*1(1+1)/2=7 we are happy :) 2. (called inductive step) Suppose this is true: 7+ 14+ 21+ 21 +28 +.....7N= 7N(N+1)/ 2 3. prove it for N+1 (we can use the inductive step) 7+ 14+ 21+ 21 +28 +.....7N+7(N+1)= this bit is 7+ 14+ 21+ 21 +28 +.....7N=7N(N+1)/ 2 so the whole is 7N(N+1)/ 2 +7(N+1)= 7(N+1)(N/2+1)=7(N+1)(N+2)/2 This completes the proof. (can you see why?)

  27. anonymous
    • 5 years ago
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    and we are very happy now :)

  28. anonymous
    • 5 years ago
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    thank you

  29. anonymous
    • 5 years ago
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    I have more sorry its confusing to me

  30. anonymous
    • 5 years ago
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    A ball bounces 400 ft every second after that its (3/4) how many times does it bounce til it completely stops

  31. anonymous
    • 5 years ago
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    no prob

  32. anonymous
    • 5 years ago
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    I figured its a geometric sequence

  33. anonymous
    • 5 years ago
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    yep it is

  34. anonymous
    • 5 years ago
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    so how do I solve?

  35. anonymous
    • 5 years ago
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    I dont understand the problem completely

  36. anonymous
    • 5 years ago
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    after that its (3/4) what does this mean?

  37. anonymous
    • 5 years ago
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    so the common ratio is 3/4 400, 300... and so on

  38. anonymous
    • 5 years ago
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    so 3/4 of 300 then the next and next.. but I need to find out how many times it bounces til it stops

  39. anonymous
    • 5 years ago
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    It will never stop than

  40. anonymous
    • 5 years ago
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    a number multiplied by 3/4 a lot of times will be a small number but will never reach 0 in maths we say it is tending to 0 as we tend to infinity

  41. anonymous
    • 5 years ago
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    well until it hits the ground?

  42. anonymous
    • 5 years ago
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    oh I am stupid, I think it asks for the use of this formula: a/(1-r)

  43. anonymous
    • 5 years ago
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    here a=400 1-r=1/4 so the result is 1600

  44. anonymous
    • 5 years ago
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    ohhh i seee :)

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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