Results 1 to 5 of 5

Math Help - Sum

  1. #1
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432

    Sum

    Hi everyone, I didn't know where to ask this, so I thought I might just ask here:

    The sum:

    1/2 + (1/3+2/3) + (1/4+2/4+3/4) + (1/5+2/5+3/5+4/5) + ... + (1/100+...+99/100) is:

    (A) 105 (B) 245 (C) 2475 (D) 3215 (E) 2635

    Could someone tell me the fastest way of being able to simplify it, and I'm not allowed to use a calculator, thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by DivideBy0 View Post
    Hi everyone, I didn't know where to ask this, so I thought I might just ask here:

    The sum:

    1/2 + (1/3+2/3) + (1/4+2/4+3/4) + (1/5+2/5+3/5+4/5) + ... + (1/100+...+99/100) is:

    (A) 105 (B) 245 (C) 2475 (D) 3215 (E) 2635

    Could someone tell me the fastest way of being able to simplify it, and I'm not allowed to use a calculator, thanks.
    Hello,

    I've attached an image of my calculations.

    EB
    Attached Thumbnails Attached Thumbnails Sum-sumofsums.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    Thanks... I only have a wikipedian-based knowledge of sums, however, so although I understand the working, I don't understand how you come to that conclusion, it was on a Grade 9-10 competition paper after all...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,751
    Thanks
    651
    Hello, DivideBy0!

    The sum:

    1/2 + (1/3 + 2/3) + (1/4 + 2/4 + 3/4) + (1/5 + 2/5 + 3/5 + 4/5) + ... + (1/100 + ... + 99/100)

    is: .(A) 105 . (B) 245 . (C) 2475 . (D) 3215 . (E) 2635

    We have: .S .= .1/2 + (1+2)/3 + (1+2+3)/4 + (1+2+3+4)/5 + ...

    . . Each numerator is the nth triangular number: .n(n + 1)

    . . Each denominator is: n + 1

    Hence, each term is: .n(n+1)/(n+1) .= .n

    . . . - . - . . . .
    99 . . . . . . . . . . . .99
    Then: .S . = . ∑ n . = . ∑ n . = . 99(100) . = . 2475
    . . . . . . . . . .
    n=1 . . . . . . . . . . .n=1

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    Ah, yes, thanks both of you, it made it so much clearer to form triangular numbers over the same denominator.

    It's easiest for me to think of it like this:
    Attached Thumbnails Attached Thumbnails Sum-sum.jpg  
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum