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Count all even from 1 to n

  1. Oct 9, 2011 #1
    Hi guys,

    I am noobie in number theory so if something exists better than this equation I did please don't bash me this is my first equation.

    Today I was thinking of a way to count all even from 1 to n,so this way I thought about is like this.

    Let me first write the equation,then I will explain the logic.


    iNumber = ((iNumber + 1) * (iNumber / 2)) - (((iNumber + 1) * (iNumber / 2) - (iNumber / 2)) / 2)


    Okay let me explain how I got this logic,so lets pick odd numbers from 1 to 10 and see how they form.
    1 = 0 + 1;
    3 = 2 + 1;
    5 = 4 + 1;
    7 = 6 + 1;
    9 = 8 + 1;

    So here we have 5 ones,so for example if we removed those ones from the a equation that you add 1 to 10 from it will be like this;

    Normal numbers which you added 1 to 10 ->

    1 + 2 + 2 + 1 + 4 + 4 + 1 + 6 + 6 + 1 + 8 + 8 + 1 + 10 -
    1 + 1 + 1 + 1 + 1,so the numbers now become

    0 + 2 + 2 + 4 + 4 + 6 + 6 + 8 + 8 + 10 = 50,but thats 2x,so if we got that number and divide by 2 it will be 25,thats how much odds add up,so if we got the total numbers from 1 to 10 and minus 25 we should get the even number we want;

    So lets put that in the equation and see if its right;

    ((11) * (5)) - (((11) * (5) - (5)) / 2) = 55 - 25 = 30;

    2 + 4 + 6 + 8 + 10 = 30;

    So the equation is right I am sure this has been done by another mathmetician,but its good to think about it the logic of it is great.
     
  2. jcsd
  3. Oct 9, 2011 #2
    Hi, genericcoder,
    I'll try to illustrate how your formula connects with other formulas known to mathematicians; in other words, how to prove (and simplify) your formula, if this is of any use to you.

    A formula well known by mathematicians (and possibly by you) is a formula for the sum of the first consecutive integers; for example, 1 + 2 + 3 + 4 + 5. The formula is[tex]\frac {k (k+1)} 2[/tex]In this example, 1 + 2 + 3 + 4 + 5 = 5 * 6 / 2 = 30 / 2 = 15.

    You want to calculate the sum of consecutive even numbers, such as 2 + 4 + 6 + 8 + 10, and that is twice the sum 1 + 2 + 3 + 4 + 5. So, if your number is 10, you would apply the above formula with 10/2 = 5 (which gives you 1+2+3+4+5; we did that and obtained 5*6/2 = 15) and then multiply the result by two to obtain 2+4+6+8+10 (= 15*2 = 30).

    So the sum of the first even numbers can be expressed as twice the above formula, that is, [itex]k(k+1)[/itex], but using a value of [itex]\displaystyle {k=\frac n 2}[/itex]:[tex]\begin{align*}\frac n 2 \left( \frac n 2 + 1 \right) &= \frac {n^2} 4 + \frac n 2 = \frac{n^2 + 2n} 4 \\ &= \frac {n(n+2)} 4 & \mbox{(eq. 1)}\end{align*}[/tex]
    In your example, if you evaluate [itex]\displaystyle {\frac {n(n+2)} 4}[/itex] with [itex]n=10[/itex], you obtain 10*12/4 = 30, as you did.

    The formula you give can be simplified to this one marked as (eq. 1), with a little algebra. Your formula is[tex]\displaystyle {(n+1)\frac n 2 - \frac {(n+1)\frac n 2 - \frac n 2} 2}[/tex]
    Notice that the numerator of the big fraction on the right is [itex]\displaystyle (n+1)\frac n 2 - \frac n 2[/itex]; in other words, (n+1)*something - something. If you have (n+1) of something, and subtract one of something, you are left with n*something. So we begin by simplifying your formula to[tex]\displaystyle {(n+1)\frac n 2 - \frac {n\frac n 2} 2}[/tex]
    or[tex](n+1)\frac n 2 - \frac {n^2} 4[/tex]
    Expanding the parentheses on the left part, we obtain[tex]\begin{align*}& \frac {n^2} 2 + \frac n 2 - \frac {n^2} 4 \\ &= \frac {n^2} 4 + \frac n 2 \\ &= \frac {n(n+2)} 4 \end{align*}[/tex]
    just as we did above for (eq. 1).

    Hope this helps!
     
  4. Oct 9, 2011 #3
    @Dodo

    Thanks alot that makes sense,also this logic for calculating the even is more efficient than mine I made it more complex.
     
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