Even integers problem

Jan 2008
70
10
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
Jul 2008
1,281
426
Reading, UK
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
Jun 2008
16,217
6,765
North Texas
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 $k > 2$ ...)

\(\displaystyle k = 6\)
 
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Feb 2015
2,255
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Ottawa Ontario
Sefah, please start your own thread;
hijacking somebody else's means nobody (usually) will look at it...
 
Sep 2015
2
0
US
How do I do that cause I'm a new user
 
Feb 2015
2,255
510
Ottawa Ontario
Click on the topic name (probably "Algebra" in your case),
then click on "start new thread"