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Is there an explicit formula for the sum of all even integers between $x$ and $y$? I don't know about one. I'm writing a program in Python and would appreciate some help.

I know that the sum of the first $n$ even integers is $n^2 + n$.

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  • $\begingroup$ Yes. And it's pretty straightforward to derive. Once you know how to add the first n numbers (and why) offsetting the starting points, having them multiple of a number like all even or all the same remainder of a multiple, are straightforward to take into accout. $\endgroup$ – fleablood Feb 15 '19 at 20:22
  • $\begingroup$ "I'm writing a program in python and would appreciate some help" Is the intention of the exercise to learn how to write a For loop correctly? If so, why are you asking this question here? Are you merely wanting a way to confirm your results? It should be clear whether or not the code was written correctly. $\endgroup$ – JMoravitz Feb 15 '19 at 20:30
  • $\begingroup$ No, I want an explicit formula. It's not an exercise, it's for a personal thing. It will make the code look nicer $\endgroup$ – user592402 Feb 15 '19 at 20:34
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Let's assume $x$ and $y$ are even, and $y > x$. Let $x = 2r$ and $y = 2s$, where $r$ and $s$ are integers. So $r = x/2$ and $s = y/2$, which we'll use later. Then you're looking for $2r + 2(r+1) + \ldots + 2s$. You have the formula for the sum of the first $n$ even integers, so you can write this as

$$ (2 + 4 + \cdots + 2s) - (2 + 4 + \cdots + 2(r-1)) $$.

Now the first thing there is the sum of the first $s$ even integers, and the second is the sum of the first $r-1$, so that's just

$$ s^2 + s - ((r-1)^2 + (r-1)) $$

or, simplifying a bit,

$$ s^2 + s - r^2 + r. $$

Remembering that $r = x/2$ and $s = y/2$, this is

$$ {y^2 \over 4} + {y \over 2} - {x^2 \over 4} + {x \over 2} $$

or, if you want everything over a common denominator,

$$ {y^2 + 2y - x^2 + 2x \over 4}. $$

For a sanity check substitute $x = 6, y = 12$. Doing the sum explicitly, you get $ 6+ 8 + 10 + 12 = 36$. From this formulas, you get

$$ {12^2 + 2\times 12 - 6^2 + 2\times 6 \over 4} = {144 + 24 - 36 + 12 \over 4} = {144 \over 4} = 36. $$

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Apply induction:

$$n^2+n+2(n+1)=(n+1)(n+2)=(n+1)^2+(n+1). $$

Correspondingly the sum of even numbers between $2m $ and $2n $ (including both) is $$ n(n+1)-m (m-1). $$

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Hint: $x+(x+2)+.....+y=$

$k*x +(0+2+4+....+(y-x))=$

$k*x+2 (0+1+2+3+....) $

Hint 2:

$x+(x+2)+...y=$

$(2+4+....+y)-(2+4+... (x-2)) $

Hint 3: $(2+4+6+......+m)+(m+......+6+4+2)=(m+2)*k$

$(x+(x+2)+...+y)+(y+(y-2)+.....+x)=(x+y)*k $

.....

There's no surprises and if an idea works in one case, the same idea will work in another.

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