A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

MATHEMATICAL INDUCTION

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[1^{2}+4^{2}+7^{2}+....+(3n-2)^{2}=\frac{ n(6n ^{2}-3n-1) }{ 2 }\]

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How would I do s(k+1)? @ganeshie8

  3. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    see here http://openstudy.com/users/irishboy123#/updates/55d3474fe4b0554d62727bc4

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay thanks that shows me the work but I dont understand it, can you explain to me step by step? @IrishBoy123

  5. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sure in inductions, you first prove it is true for some value of n. so establish that bit first.

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I already did that part: n=1 and n=k I just dont know how to d the n=k+1

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Honestly, I don't get it.

  8. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    OK, one important thing. last time i helped on this the OP specified "for all positive integers n" so yes you had to start with n = 1. we will use induction to prove that it must therefore be true for n = 2, and thuse n = 3 and thus .... so the next step is to assume it is true for n = k. that means the following is assumed true |dw:1440079974650:dw|

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah I have that

  10. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i follows therefore that : |dw:1440080053278:dw| that has to be true as we have added the same amount to each side as it happens what we added to LHS was next term in the series

  11. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and the critical step, do you know that too?

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that step just confused me im trying to understand it just a min

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Question: is it not that for n =k, we have \(....+(3k-2)^2= \dfrac{k(6k^2-3k-1)}{2}\)

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1440080370055:dw|

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Woah im so lost now

  16. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1440080425206:dw|

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh okay so 3(k+1-2)^2 takes the place of 6k^2....

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah

  19. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    slow down a bit gives me a minute to summarise where we are: will take less than a minute

  20. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1440080585980:dw|

  21. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1440080618737:dw|

  22. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so the final step is to show that the RHS is as the original equation would predict ie |dw:1440080671229:dw|

  23. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and that's proof done, if we complete that final step

  24. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so we now wrestle with the algebra trying to show these 2 are in fact equal.

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I did, they are not the same :(

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I show you

  27. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    1 Attachment
  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright I have to go ill be back though sorry You guys can continue helping crazyand... I will catch up with you guys

  29. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that's all you need to know, the rest is the algebra. it's turgid, that's why i shared my working -- which shows one approach

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \(\dfrac{k(6k^2-3k-1)}{2}+(3(k+1)-2)^2\\=\dfrac{k(6k^2-3k-1)}{2}+(3k+1)^2\\=\dfrac{k(6k^2-3k-1)}{2}+9k^2+6k+1\\=\dfrac{k(6k^2-3k-1)+18k^2+12k+2)}{2}\\\) \(=\dfrac{6k^3-3k^2-k+18k^2+12k+2}{2}\) \(=\dfrac{6k^3+15k^2+11k+2}{2}\)

  31. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    right, so far. at this point i divided by k + 1 as we know that is on the RHS if the statement is correct

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    while other is \(\dfrac{(k+1)(6(k+1)^2-3(k+1)-1}{2}\) \(=\dfrac{(k+1)(6(k^2+2k+1))-3k-3-1}{2}\) \(\dfrac{(k+1)(6k^2+12k+6)-3k-4}{2}\) \(=\dfrac{6k^3+12k^2+6k+6k^2+12k+6-3k-4}{2}\) \(=\dfrac{6k^3+18k^2+15k+2}{2}\) Which is different from the previous one. :(

  33. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    check brackets

    1 Attachment
  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    YYYYYYYYYYYES!! I got you. Finally. Thanks for being patient to me.

  35. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thank you for showing me a shorter proof. i don't know what the hell i was doing using long division last time i did this with someone .... just plain nuts.

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    We learned from each other, right? hehehe..

  37. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.