# Even integers problem

#### donnagirl

If the sum of the even integers between 1 and k inclusive is equal to 2k, what is the value of k?

How do you prove that k must be 6 and not larger or smaller than 6?

#### Matt Westwood

MHF Hall of Honor
You can make a start by putting together an expression for the sum of the even integers between 1 and k inclusive.

Let S be the sum of the even integers between 1 and k inclusive.

Then:
$S = 2 + 4 + 6 + \cdots + (k-2) + k$

As k is even, you can say r = 2k and so:

$S = 2 (1 + 2 + 3 + \cdots + (r-1) + r)$

Now the sum of the first $r$ numbers is well-known, it's the $r$th triangle number and we have:

$1 + 2 + 3 + \cdots + (r-1) + r = \dfrac {r (r+1)} 2$

Now we can keep it simple and say $2k = 4r$ and so:

$S = 2 (1 + 2 + 3 + \cdots + (r-1) + r) = 4r = 2 \dfrac {r (r+1)} 2 = r (r+1)$

So you can build a quadratic in $r$ and so get $k$.

#### skeeter

MHF Helper
If the sum of the even integers between 1 and k inclusive is equal to 2k, what is the value of k?

How do you prove that k must be 6 and not larger or smaller than 6?
inclusive? 1 is not an even integer.

note the sum of the first "$n$" even integers is ...

$2+4+6+...+2n=n(n+1)$

in this case, $k = 2n \implies n = \frac{k}{2}$, so the sum of the even integers between 2 and k inclusive is $\displaystyle \frac{k}{2}\left(\frac{k}{2}+1\right) = 2k$

solving for $k$ ...

$$\displaystyle \frac{k^2}{4} + \frac{k}{2} = 2k$$

$$\displaystyle k^2 + 2k = 8k$$

$$\displaystyle k^2 - 6k = 0$$

$$\displaystyle k(k - 6) = 0$$

$k = 0$ (which can't be because $k > 2$ ...)

$$\displaystyle k = 6$$

1 person

#### DenisB

hijacking somebody else's means nobody (usually) will look at it...

#### sefah76

How do I do that cause I'm a new user

#### DenisB

Click on the topic name (probably "Algebra" in your case),
then click on "start new thread"