Results 1 to 7 of 7

Thread: sum of first 1000 odd numbers is 100,000,000 but what is the

  1. #1
    Guest

    sum of first 1000 odd numbers is 100,000,000 but what is the

    I'm trying to determine the formula for the sum of the first 1000 odd digits.
    I know that the sum of all number from 1 to 1000 is n(n+1)/2 but I can't figure out the odd number formula. Please help!

  2. #2
    Junior Member
    Join Date
    Jul 2005
    Location
    Davis, CA
    Posts
    187

  3. #3
    Senior Member
    Join Date
    Oct 2003
    Location
    Wisconsin
    Posts
    1,917
    Another way.
    If you know that then:
    The first number in an arithmetic progression is a and the difference is d :
    y=a+(a+d)+(a+2d)+(a+3d)...+(a+(n-1)d) =
    n*a+sum(from 1 to (n-1))*d =
    n*a+ (d*(n-1)*n/2) =
    (2a + d*(n-1))*n/2
    I hope this helps. If you need more, come back with a post-reply.
    You are better off putting problems in the general areas. More people are avauilable to answer them there.
    Gene

  4. #4
    Full Member
    Join Date
    Jul 2005
    Location
    Long Island, NY
    Posts
    876
    Surprisingly, the sum of the first n odd integers is n^2.

    How do you find the sum of a series of odd or even numbers?

    Consider the odd numbers.

    n......1......2......3......4......5......6
    N.....1......3......5......7......9.....11
    Sum.1.....4......9.....16.....25....35 (Look familiar?)
    Diff......3......5......7......9......11
    Diff.........2.......2......2......2

    With the 2nd differences constant, the sequence of sums forms a fintite difference sequence.
    The general expression for the nth term (the sum of n odd numbers) is of the form an^2 + bn + c.
    Using the given data, we can write

    a(1^2) + b(1) + c = 1
    a(2^2) + b(2) + c = 4
    a(3^2) + b(3) + c = 9

    Solving, a = 1, b = 0 and c = 0.
    Therefore, the general expression for the sum of "n" consecutive odd numbers starting with 1 is S(n) = n^2.

    Consider the even numbers.

    n......1......2......3......4......5......6
    N.....2......4......6......8.....10....12
    Sum2......6.....12....20.....30....42
    Diff.....4.......6......8.....10.....12
    Diff..........2......2......2......2

    With the 2nd differences constant, the sequence of sums forms a fintite difference sequence.
    The general expression for the nth term (the sum of n odd numbers) is of the form an^2 + bn + c.
    Using the given data, we can write

    a(1^2) + b(1) + c = 2
    a(2^2) + b(2) + c = 6
    a(3^2) + b(3) + c = 12

    Solving, a = 1, b = 1 and c = 0.

    Therefore, the general expression for the sum of "n" consecutive even numbers starting with 2 is S(n) = n^2 + n = n(n + 1).
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  5. #5
    Full Member
    Join Date
    Jul 2005
    Location
    Long Island, NY
    Posts
    876
    The sum of the first n odd numbers is n^2

    n....1....2.....3.....4
    ......1 = 1
    ......1 + 3 = 4
    ......1 + 3 + 5 = 9
    ......1 + 3 + 5 + 7 + 9 = 25, etc.

    * The nth square is equal to n^2.
    n....1....2....3....4....5....6....7....8....9.... 10
    Sq..1...4....9....16..25..36..49..64...81...100

    The sum of the first 1000 odd numbers is 1000^2 = 1,000,000.
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  6. #6
    Elite Member
    Join Date
    Jan 2005
    Location
    Lexington, MA
    Posts
    5,607

    Re: sum of first 1000 odd numbers is 100,000,000 but what is

    Hello, aciokajlo!

    If your brain isn't already overloaded, here another approach.
    . . It's primitive and quite long, but it's nice to have options, isn't it?

    ]I'm trying to determine the formula for the sum of the first 1000 odd digits.
    I know that the sum of all number from 1 to 1000 is n(n+1)/2
    but I can't figure out the odd number formula.
    You already know the sum of all the numbers from 1 to 1000, right?
    . . . (1000)(1001)/2 .= .500,500

    Now find the sum of the even numbers from 2 to 1000.
    . . . 2 + 4 + 6 + ... + 1000 .= .2(1 + 2 + 3 + ... + 500)

    This is: . 2 x (sum of 1 to 500) .= .2 x (500)(501)/2 .= .250,500


    Therefore: .(sum of odd) .= .(sum of all) - (sum of even) .= .500,500 - 250,500 .= .250,000


    . . See? . . . You <u>could</u> have figured it out yourself!
    I'm the other of the two guys who "do" homework.

  7. #7
    Full Member
    Join Date
    Jul 2005
    Location
    Long Island, NY
    Posts
    876
    To repeat:

    Surprisingly, the sum of the first n odd integers is simply n^2.

    The first odd number is 1. The 1000th odd number is 2(1000) - 1 = 1999.

    How do you find the sum of a series of odd numbers?

    Consider the odd numbers.

    n..........1......2......3......4......5......6... ...7......8......9......10.............n
    Odd#...1......3......5......7......9.....11.....13 .....15....17.....19 .........2n-1
    Sum.....1.....4......9.....16.....25....36........ ....................................n^2

    Notice anything? The sum of any consecutive sequence of odd numbers starting with 1 is n^2.

    I gave you the analytical method earlier.
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

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
  •