1. ## A simple proof?

Prove that:
2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)
When n > 0

Getting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though...

2. Use induction to prove. Show the formula holds for the base case, assume it hold for the $\displaystyle n^{th}$ case and derive the $\displaystyle (n+1)^{th}$ case.

Why don't you try and let me know where you get stuck.

3. Or wait!

$\displaystyle 2 + 4 + 6 + 8 ... + 2(n-1) = 2(1+2+3+...+(n-1)) = 2(\frac{n(n-1)}{2}) = n(n-1)$

You may know the formula to obtain the sum of the numbers from $\displaystyle 1$ to $\displaystyle n.$ It is $\displaystyle \frac{n(n+1)}{2}.$

Here is an example:
Sum the numbers $\displaystyle 1$ to $\displaystyle 5.$ Then,

$\displaystyle 1+2+3+4+5 = \frac{5\times 6}{2} = 15.$

4. Hm. Gimme a second to type as I'm thinking.

n = 1
2(1-1) = 1 * (1-1) = 0

n = k+1
2 + 4 + 6 ... + 2k = k^2 + k
2 + 4 + 6 ... + k = k^2
2 + 4 + 6 ... + 2(k-1) = k^2 - k
2 + 4 + 6 ... + 2(k-1) = k * (k-1)

Oh, wow. Thanks so much!

EDIT: Or, that second post works too. Heh, either one. Thanks for the two methods, at least.

5. Originally Posted by BlackBlaze
Prove that:
2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)
When n > 0

LHS:

$\displaystyle 2 + 4 + 6 + 8 +\dots+ 2(n-4)+ 2(n-3) + 2(n-2) + 2(n-1) = S$

Swapping the order of the LHS

$\displaystyle 2(n-1) + 2(n-2)+ 2(n-3) + 2(n-4) +\dots+ 8 + 6+ 4 +2= S$

$\displaystyle \underbrace{2n+2n + 2n+\dots+ 2n}_{\text{n-1 times}} = 2S$

$\displaystyle 2n(n-1) = 2S$

$\displaystyle n(n-1) = S$

$\displaystyle S = n(n-1)$

6. Hello,
here's a proof with the sum symbol (looks cool ).
What you are trying to prove is that $\displaystyle \sum_{k=1}^{n - 1} 2k = n(n - 1)$. Let's work with this :

$\displaystyle \sum_{k=1}^{n - 1} 2k$

Note that each term is even, we can factorize ! ($\displaystyle 2 + 4 + 6 = 2(1 + 2 + 3)$). So :

$\displaystyle \sum_{k=1}^{n - 1} 2k = 2 \left ( \sum_{k=1}^{n - 1} k \right )$

Now you may use the fact that $\displaystyle \sum_{k=1}^{n} k = \frac{1}{2}n(n + 1)$, and your result follows. This is a theorem but it can be proven :

$\displaystyle \sum_{k=1}^{n} k = 1 + 2 + 3 + \cdots + k$

1. Assume $\displaystyle n$ is even.

Rearranging the terms (putting brackets for better comprehension). This is valid because $\displaystyle n$ is even :

$\displaystyle \sum_{k=1}^{n} k = \left [ 1 + \left ( n - 1 \right ) \right ] + \left [ 2 + \left ( n - 2 \right ) \right ] + \cdots + \frac{n}{2} + n$

The dotted part stops when we reach $\displaystyle n - \frac{n}{2}$ (excluded) because we cannot include that term twice. (*)

This can be simplified :

$\displaystyle \sum_{k=1}^{n} k = n + n + \cdots + \frac{n}{2} + n$

Using fact (*), we simplify by multiplication :

$\displaystyle \sum_{k=1}^{n} k = \left ( \frac{n}{2} - 1 \right ) n + \frac{n}{2} + n$

Expanding and simplifying :

$\displaystyle \sum_{k=1}^{n} k = \frac{n^2}{2} - n + \frac{n}{2} + n = \frac{n^2}{2} + \frac{n}{2} = \frac{n^2 + n}{2} = \frac{n(n + 1)}{2} = \frac{1}{2} n(n + 1)$

2. Assume $\displaystyle n$ is odd.

Rearranging the terms (putting brackets for better comprehension). This is valid because $\displaystyle n$ is odd :

$\displaystyle \sum_{k=1}^{n} k = \left [ 1 + n \right ] + \left [ 2 + \left ( n - 1 \right ) \right ] + \cdots + \frac{n + 1}{2}$

The dotted part stops when we reach $\displaystyle n - \frac{n + 1}{2}$ (excluded) because if we included it we would include two terms twice! (*)

This can be simplified :

$\displaystyle \sum_{k=1}^{n} k = (n + 1) + (n + 1) + \cdots + \frac{n + 1}{2}$

Using fact (*), we simplify by multiplication :

$\displaystyle \sum_{k=1}^{n} k = \left ( \frac{n + 1}{2} - 1 \right )(n + 1) + \frac{n + 1}{2}$

Expanding and simplifying (using the trick that $\displaystyle \frac{n + 1}{2} - 1 = \frac{n - 1}{2}$) :

$\displaystyle \sum_{k=1}^{n} k = \left ( \frac{n + 1}{2} - 1 \right )(n + 1) + \frac{n + 1}{2} = \frac{(n - 1)(n + 1)}{2} + \frac{n + 1}{2} =$$\displaystyle \ \frac{(n - 1)(n + 1) + (n + 1)}{2} = \frac{(n + 1)(n - 1 + 1)}{2} = \frac{1}{2} n(n + 1)$

Using the theorem we just proved (the proof is correct, just a bit messed up, needs better editing), we get that :

$\displaystyle \sum_{k=1}^{n - 1} 2k = 2 \sum_{k=1}^{n - 1} k = 2 \times \left ( \frac{1}{2} (n - 1)(n - 1 + 1) \right ) = n(n - 1)$

And we can conclude.

Does it make sense ?

EDIT : I just saw Pickslides' proof. Mine seems a bit redundant now

7. Originally Posted by BlackBlaze
Prove that:
2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)
When n > 0

Getting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though...
Well, the sum of n-1 number is n-1 times their average. Since this is an arithmetic sequence, the average of all the numbers is just the average of the smallest and largest: $\displaystyle \frac{2+ 2(n-1)}{2}= 1+ (n- 1)= n$.