Results 1 to 3 of 3

Thread: Sum of n terms of sequence: a_1 sqrt[a_1] + a_2 sqrt[a_2] + ... + a_n sqrt[a_n]

  1. #1
    New Member
    Join Date
    Sep 2017
    Posts
    2

    Sum of n terms of sequence: a_1 sqrt[a_1] + a_2 sqrt[a_2] + ... + a_n sqrt[a_n]

    Hi, I have an exercise that would be considered to belong to calculus, but i got stuck on solving a sum(even further, i have the final step at the answer, but I don't understand how they got to the conclusion).
    Basically i have this sum, but i don't know the steps between where i got stuck and the last step from the answer.

    Where I got stuck:

    . . . . .[tex]a_1\, \sqrt{\strut a_1\,}\, +\, a_2\, \sqrt{\strut a_2\, }\, +\, ...\, +\, a_n\, \sqrt{\strut a_n\,}\, =[/tex]

    Last step from the answer:

    . . . . .[tex]a_1\, -\, a_2\, +\, a_2\, -\, a_3\, +\, ...\, +\, a_n\, -\, a_{n+1}\, =[/tex]

    . . . . .[tex]a_1\, -\, a_n[/tex]

    I wold really appreciate if someone could explain me the steps in between.
    Attached Images Attached Images
    Last edited by stapel; 09-06-2017 at 02:54 PM. Reason: Typing out the text in the graphic.

  2. #2
    Elite Member
    Join Date
    Jun 2007
    Posts
    17,577
    Quote Originally Posted by Cezar View Post
    Hi, I have an exercise that would be considered to belong to calculus, but i got stuck on solving a sum(even further, i have the final step at the answer, but I don't understand how they got to the conclusion).
    Basically i have this sum, but i don't know the steps between where i got stuck and the last step from the answer.

    Where I got stuck:

    . . . . .[tex]a_1\, \sqrt{\strut a_1\,}\, +\, a_2\, \sqrt{\strut a_2\, }\, +\, ...\, +\, a_n\, \sqrt{\strut a_n\,}\, =[/tex]

    Last step from the answer:

    . . . . .[tex]a_1\, -\, a_2\, +\, a_2\, -\, a_3\, +\, ...\, +\, a_n\, -\, a_{n+1}\, =[/tex]

    . . . . .[tex]a_1\, -\, a_n[/tex]

    I wold really appreciate if someone could explain me the steps in between.
    Did they provide any relationship between a1 and a2 and a3 ..... an?
    Last edited by stapel; 09-06-2017 at 02:55 PM. Reason: Copying typed-out graphical content into reply.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    New Member
    Join Date
    Sep 2017
    Posts
    2
    Quote Originally Posted by Subhotosh Khan View Post
    Did they provide any relationship between a1 and a2 and a3 ..... an?
    Thank you for response. Apparently I've done the exercise with another formula that I was supposed to do. Basically when I was supposed to study the monotony of the sequence I used formula 1) instead of 2), and I haven't seen the relation from the second formula.



    1) If [tex]\dfrac{a_{n+1}}{a_n}\, \geq\, 1,[/tex] then the sequence is increasing; otherwise, it is decreasing.

    2) If [tex]a_{n+1}\, -\, a_n\, \geq\, 0,[/tex] then the sequence is increasing; otherwise, it is decreasing.

    And, solving with the second formula, knowing that

    . . .[tex]a_{n+1}\, =\, a_n\, \left(1\, -\, \sqrt{\strut a_n\,}\right)[/tex]

    ...we get:

    . . .[tex]a_{n+1}\, -\, a_n\, =\, a_n\, \left(1\, -\, a_n\right)\, -\, a_n\, =\, -a_n\, \sqrt{\strut a_n\,}[/tex]

    Further:

    . . .[tex]a_{n+1}\, -\, a_n\, =\,-a_n\, \sqrt{\strut a_n\,}\, \big|\, \cdot\, -1[/tex]

    . . .[tex]a_n\, \sqrt{\strut a_n\,}\, =\, a_n\, -\, a_{n+1}[/tex]

    And that would be the formula.



    Thank you for your time, I guess next time I'll try all formulas before asking
    Attached Images Attached Images
    Last edited by stapel; 09-06-2017 at 03:03 PM.

Tags for this Thread

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •