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Rules of consequent addition

  1. Jun 16, 2015 #1
    I have recently discovered some mathematical rules by trial and error. Some of these may be already known to people of mathematics.
    1. This one is common: Sum of all integers 1+2+3+...=0.5(n^2 + n)
    2. Sum of odd integers: 1+3+5+7+9......= n^2
    3. Sum of even integers: 2+4+6+8.....= n^2 + n eg: Sum of first two: 2+4= 2^2 +2 =6
    4. Sum of gap odd integers(odd integers with one gap) : 1+5+9+13+17....= (n+1)(n+2) eg. 1+5=(1+1)(1+2)=6
    5.a. Sum of gap even integers: 2+6+10+14....= 2n^2 eg 2+6+10= 2(3)^2=18
    5.b. Sum of gap even integers: 4+8+12+16...= 4(1+2+3+4+...) eg:4+8+12= 4(1+2+3)= 24
     
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  3. Jun 16, 2015 #2

    PeroK

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  4. Jun 16, 2015 #3

    phinds

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    The algebraic derivations of all these is trivial, so yes, they are known.
     
  5. Jun 16, 2015 #4

    Stephen Tashi

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    If you want to develop such rules in a systematic fashion, try starting with a function like F(n) = n^2 and forming the series give by its differences: (F(1) - F(0)) + ( F(2) - F(1)) + (F(3) - F(2)) + ....(F(n+1) - F(n)). This series sums to F(n+1) - F(0).

    For functions that are simple to write down, the "answer" F(n+1) - F(0) looks simple while the series whose terms are T(i) = F(i+1) - F(i) can look complicated.
     
  6. Jun 17, 2015 #5
    Sum of odd integers with one gap = n(2n-1)
     
  7. Jun 17, 2015 #6

    micromass

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    Try to find a general formula now which includes all those cases:

    [tex]a+ (a+b) + (a + 2b) + (a + 3b) + ... + (a + nb)[/tex]

    And try to find the sum of squares:

    [tex]1 + 4 + 9 + ... + n^2[/tex]

    Or if you're up for a challenge, the more general

    [tex]1 + 2^k + 3^k + ... + n^k[/tex]
     
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