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anonymous

  • 5 years ago

Use mathematical induction prove that: ¼+2*¼+4*¼+....+[2(n-1)+¼]=4n²-3n/4 these are sooo confusing

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  1. anonymous
    • 5 years ago
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    you've got to take steps: 1) Show that P(1) is true 2) Assume P(k) is true 3) find P(k+1) ^_^

  2. anonymous
    • 5 years ago
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    now to show that p(1) is true : P(1) : 2(1-1) = 4(1)^2-3(1)/4<-- are you sure of the equation?

  3. anonymous
    • 5 years ago
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    theres a 1/4 lol

  4. anonymous
    • 5 years ago
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    oh right >_< thank you :)

  5. anonymous
    • 5 years ago
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    so after that, you do the calculation and if it's true proceed to step (2)

  6. anonymous
    • 5 years ago
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    its the same as the other induction question you put essentually

  7. anonymous
    • 5 years ago
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    expect this time it wont be so easy to simplify the expression in step 3

  8. anonymous
    • 5 years ago
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    2) Assume p(k) :<-- in this case, replace the n's with k ^_^

  9. anonymous
    • 5 years ago
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    except

  10. anonymous
    • 5 years ago
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    are you getting the picture 2BGood? :)

  11. anonymous
    • 5 years ago
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    i know it's like the other one, and you explained the other one so well, but i look at the problem and go????????????

  12. anonymous
    • 5 years ago
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    ksajflkdaf

  13. anonymous
    • 5 years ago
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    yep ^_^ , just following the steps dear

  14. anonymous
    • 5 years ago
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    lol Mak :)

  15. anonymous
    • 5 years ago
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    follow*

  16. anonymous
    • 5 years ago
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    Assume S(k)= ( 4k^2 -3k) / 4 We need to prove that S(k+1) = [4(k+1)^2 -3(k+1) ] / 4

  17. anonymous
    • 5 years ago
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    alright, elec got it from here, good luck :)

  18. anonymous
    • 5 years ago
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    remember that I got the second line above by sub n=k+1 into the expression on the RHS of the sum

  19. anonymous
    • 5 years ago
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    Now, we also know the result S(k+1) = S(k) +T(k+1) , very important result you need to know to summation induction questions

  20. anonymous
    • 5 years ago
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    all you've got to do is plug in k+1 in the sequence then prove that the LHS and RHS are equal. LHS= .... (simplify it) then check if you get the same thing with the RHS >_< LOL

  21. anonymous
    • 5 years ago
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    so we know that S(k+1) =[ [ 4k^2 -3k ] / 4] +( 2k + (1/4) )

  22. anonymous
    • 5 years ago
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    the first fraction comes from the assumption , the second bracket comes from subbing n=k+1 into the general term on the LHS , thats our (k+1)th term

  23. anonymous
    • 5 years ago
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    now its just a matter of algebra

  24. anonymous
    • 5 years ago
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    S(k+1) =[ 4k^2 -3k +8k +1 ] / 4 by getting over a common denominator

  25. anonymous
    • 5 years ago
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    2BGood, are you getting the picture? ^_^

  26. anonymous
    • 5 years ago
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    now , this is the exact same as what were asked to prove

  27. anonymous
    • 5 years ago
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    if you go up to the post where I made the assumption step , and expanded out the S(k+1) I wrote there, then you will get the same thing

  28. anonymous
    • 5 years ago
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    all summmation induction questions are relatively straight forward, just S(k+1)= S(k) + T(k+1) , know thay and you are good

  29. anonymous
    • 5 years ago
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    divisability induction questions are also relatively straight forward, but inequality induction questions are when it starts to get a bit tricky

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