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Math Help - Please help

  1. #1
    Junior Member
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    Please help

    trying to prove that the sum of consecutive odd numbers is always the difference of two sqaures.

    First I listed the odd numbers

    1; 3; 5; 7; 9,

    Then difference of two squares

    4 -1 = 3

    9-4 = 5

    16 - 4 = 12 = 5 +7 (sum of odd numbers)

    25 - 4 = 21 = 5 + 7 + 9

    How do I generalise? please help.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by fair_lady0072002 View Post
    trying to prove that the sum of consecutive odd numbers is always the difference of two sqaures.

    First I listed the odd numbers

    1; 3; 5; 7; 9,

    Then difference of two squares

    4 -1 = 3

    9-4 = 5

    16 - 4 = 12 = 5 +7 (sum of odd numbers)

    25 - 4 = 21 = 5 + 7 + 9

    How do I generalise? please help.
    Your prompt says you are "trying to prove that the sum of consecutive odd numbers is always the difference of two sqaures." However, what you are actually trying to prove is that the difference of two squares is the sum of consecutive odd numbers. I just want it to be clear you have this written backwards.

    To "prove" the sum of consecutive odd numbers is always a difference of two squares, you would have to show that:
    (2n + 1) + (2n + 3) = a^2 - b^2

    is true for any value of n = 0, 1, 2,..., where a and b are integer numbers. But this isn't true.

    For example, if 2n + 1 = 1 and 2n + 3 = 3 (where n = 0), then we have:
    1 + 3 = a^2 - b^2 --> 4 = a^2 - b^2

    but the difference of no two perfect squares is equal to 4.

    Therefore, what you are actually trying to prove is that:
    a^2 - b^2 = (2n + 1) + (2n + 3) + ... + (2n + (2m + 1))

    where m is some integer: m = 0, 1, 2,...

    How you would go about proving this, I'm not sure.
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  3. #3
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    Hello, fair_lady!

    Prove that the sum of consecutive odd numbers is always the difference of two sqaures.
    Here's a hint . . . maybe you can exploit it.

    Fact: Consecutive odd numbers are the differences of consecutive squares.
    Code:
    0  1  2  3   4   5    6   7    8
    0   1   4   9   16   25    36    49    64
     \ / \ / \ / \ /  \ /  \  /  \  /  \  /
      1   3   5   7    9    11    13    15

    So that a set of consecutive odd numbers, say, 7 + 9 + 11 + 13,
    . . will come from this arrangement:
    Code:
    0  1  2  3   4   5    6   7    8
    0   1   4   9   16   25    36    49    64
                 \                  /
                  7 +  9  + 11 +  13
    So that (7 + 9 + 11 + 13) is the difference of 7 and 3.


    Now if we can express all that in general, we'd have our proof.

    But I haven't done it yet . . . maybe you'll have better luck.

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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by fair_lady0072002 View Post
    trying to prove that the sum of consecutive odd numbers is always the difference of two sqaures.

    First I listed the odd numbers

    1; 3; 5; 7; 9,

    Then difference of two squares

    4 -1 = 3

    9-4 = 5

    16 - 4 = 12 = 5 +7 (sum of odd numbers)

    25 - 4 = 21 = 5 + 7 + 9

    How do I generalise? please help.
    The sum of m consecutive odd numbers starting from 2n+1 may be written:

    S_{n,m} = (2n+1) + (2(n+1)+1) + ... + (2(n+m-1)+1)

    ............= sum_{r=n to (n+m-1)} (2r+1) = m(2n+m)

    Thus if m is even S_{n,m} is the product of two even numbers, and if m
    is odd S_{n,m} is the product of two odd numbers.

    Let a and b be any pair of even numbers (a>=b), then write:

    a = u+v

    and

    b = u-v

    then u=(a+b)/2 and v=a - (a+b)/2 are both integers and:

    a.b=(u+v)(u-v)=u^2-v^2.

    Let a and b be any pair of odd numbers (a>=b), then write:

    a = u+v

    and

    b = u-v

    then u=(a+b)/2 and v=a - (a+b)/2 are both integers and:

    a.b=(u+v)(u-v)=u^2-v^2.


    Hence as S_{n,m} is the product of two odd numbers or of two even
    numbers it can be written as the difference of two squares (one of which
    may be 0).

    RonL
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  5. #5
    Junior Member
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    Thank you all

    I really appreciate your help
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