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ParthKohli

  • one year ago

As promised. Many of you may have solved this, but anyway.

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  1. ParthKohli
    • one year ago
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    A number sequence \(a_1, a_2, \cdots, a_n\) is such that\[a_1 =0\]\[|a_2 | = |a_1 + 1|\]\[\vdots\]\[|a_n| = |a_{n-1} + 1|\]Prove that the arithmetic-mean of \(a_1, a_2, \cdots, a_n \) is \(\ge -1/2\).

  2. ParthKohli
    • one year ago
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    @ganeshie8

  3. ParthKohli
    • one year ago
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    @imqwerty @mathmath333 @perl

  4. ganeshie8
    • one year ago
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    don't we simply get a sequence of whole numbers ?

  5. ParthKohli
    • one year ago
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    Yes, of course.

  6. ParthKohli
    • one year ago
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    Of integers.

  7. ParthKohli
    • one year ago
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    \(a_2\) can also be -1, for instance.

  8. ganeshie8
    • one year ago
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    Oh I see it.

  9. ParthKohli
    • one year ago
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    A hint is that you don't want the absolute value there. What do you do?

  10. ganeshie8
    • one year ago
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    Oh parth you give hints too early

  11. ParthKohli
    • one year ago
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    Sorry, I'm an impatient guy. :|

  12. ganeshie8
    • one year ago
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    \[\sum\limits_{i=1}^{n} {a_n}^2~~=~~\sum\limits_{i=1}^{n} (a_{n-1}+1)^2\]

  13. ParthKohli
    • one year ago
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    :D

  14. ganeshie8
    • one year ago
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    should i try telescoping ?

  15. ParthKohli
    • one year ago
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    You were at least able to extract meaning out of that hint.

  16. ParthKohli
    • one year ago
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    Nah, just expand.

  17. ganeshie8
    • one year ago
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    \[\sum\limits_{i=1}^{n} {a_n}^2-{a_{n-1}}^2~~=~~\sum\limits_{i=1}^{n} 2a_{n-1}+1\] ?

  18. ganeshie8
    • one year ago
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    idk what to do next but if i had pen and paper wid me i would try and see if it telescopes..

  19. ParthKohli
    • one year ago
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    Yeah, things cancel out.

  20. ParthKohli
    • one year ago
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    The right-side is very important.

  21. ganeshie8
    • one year ago
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    let me fix the typoes in indices \[\sum\limits_{i=\color{Red}{2}}^{n} {a_\color{Red}{i}}^2-{a_{\color{Red}{i}-1}}^2~~=~~\sum\limits_{i=\color{red}{2}}^{n} 2a_{\color{red}{i}-1}+1\]

  22. ParthKohli
    • one year ago
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    Yeah, it's mostly about the RHS now.

  23. ganeshie8
    • one year ago
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    I think i need to zero the index on right hand side

  24. ParthKohli
    • one year ago
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    Make both the upper limits \(n+1\).

  25. ganeshie8
    • one year ago
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    \[{a_n}^2-{a_1}^2~~=~~(n-1)-2a_n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}} \]

  26. ganeshie8
    • one year ago
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    looks completing the square will do the job

  27. ParthKohli
    • one year ago
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    Yeah, you can also use \(a_1 = 0\)

  28. ganeshie8
    • one year ago
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    \[({a_n}+1)^2~~=~~{a_1}^2+n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}}\]

  29. ganeshie8
    • one year ago
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    \[0~~\le ~~n+2\sum\limits_{i=\color{red}{1}}^{n} a_{\color{red}{i}}\]

  30. ganeshie8
    • one year ago
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    Yaay!

  31. ParthKohli
    • one year ago
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    Great, well done :)

  32. ganeshie8
    • one year ago
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    ty ty :)

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