A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Mathsadness

  • 5 years ago

sum{k=1 to 30} (6k^2+1) How would you find the value of this sum?

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

    sum as \[6\sum_1^{30}k^2 +\sum_1^{30}1\]

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

    second term is just 30

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do you know the formula for \[\sum_{k=1}^n k^2\]?

  4. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    its is \[\frac{n(n+1)(2n+1)}{6}\]

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

    replace n by 30, compute and you will have the answer. i will write it if you like

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

    \[6\times \frac{30(31)(71)}{6} +30\] \[30(31)(71)+30\]

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i get 66060

  9. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sum_{k=1}^{n} k^3 \] What is the formula for this??

  10. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ...

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that one is easy to remember.

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

    do you know the one for \[\sum_{k=1}^{n}k\]?

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

    it is \[\frac{n(n+1)}{2}\]

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

    the one for \[\sum_{k=1}^n k^3 \] is the square of that one. it is\[(\frac{n(n+1)}{2})^2\]

  15. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so the square of Gauss' formula??

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if that thing is called gauss's formula then yes

  17. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for the summation of n positive integers....

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

    yes. here they are, more than you want to know http://polysum.tripod.com/

  19. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wow!

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yes, i see that it is called gauss' formula. i learn more here than in skule

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if you like i can show you how to derive them, but you probably don't need to know it

  22. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how do you do that?

  23. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :P

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok we start with the first one. "gauss" and the method extends. you want \[\sum_{k=1}^n k\]

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

    write this \[\sum_{k=1}^n k^2-(k-1)^2\]

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

    it might not be obvious, but the part after the sigma is one term minus the previous one. so this is a telescoping sum, and when you are done (it will be obvious if you write out some terms) everything will be killed off except the \[n^2\]

  27. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so by observation we have \[\sum_{k=1}^n k^2-(k-1)^2=n^2\]

  28. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    then do the algebra: \[k^2-(k-1)^2 = k^2-(k^2-2k+1)=2k-1\]

  29. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    in other words there was no square in it. now we know that \[\sum_{k=1}^n 2k-1=n^2\] because the part after the sum was the same. this is equal to \[2\sum_{k=1}^nk-\sum_{k=1}^n =n^2\]

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

    now solve for \[\sum_{k=1}^nk\] using algebra and you get \[\sum_{k=1}^n k = \frac{n(n+1)}{2}\]

  31. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f course you have probably seen this proved a different way. but now the fun begins. if you want the formal for \[\sum_{k=1}^n k^2 \] you write \[\sum_{k=1}^n k^3-(k-1)^3=n^3\]

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

    do the algebra and the \[k^3\] will drop ouit (just like the k^2 did before) and you will get a summation only involving \[k^2\] and k and 1. you already have the one for k, so solve for \[\sum_{k=1}^nk^2\] and you will get your answer. this trick works all the way up, but the algebra gets really really annoying so best just to have a list

  33. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cool!

  34. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    problem 9 - 12 and and 20 - 23 work it out (not the algebra)

  35. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    have fun

  36. Mathsadness
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thx

  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.