Use induction to prove. Show the formula holds for the base case, assume it hold for the case and derive the case.
Why don't you try and let me know where you get stuck.
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...
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.
Hello,
here's a proof with the sum symbol (looks cool ).
What you are trying to prove is that . Let's work with this :
Note that each term is even, we can factorize ! ( ). So :
Now you may use the fact that , and your result follows. This is a theorem but it can be proven :
1. Assume is even.
Rearranging the terms (putting brackets for better comprehension). This is valid because is even :
The dotted part stops when we reach (excluded) because we cannot include that term twice. (*)
This can be simplified :
Using fact (*), we simplify by multiplication :
Expanding and simplifying :
2. Assume is odd.
Rearranging the terms (putting brackets for better comprehension). This is valid because is odd :
The dotted part stops when we reach (excluded) because if we included it we would include two terms twice! (*)
This can be simplified :
Using fact (*), we simplify by multiplication :
Expanding and simplifying (using the trick that ) :
Using the theorem we just proved (the proof is correct, just a bit messed up, needs better editing), we get that :
And we can conclude.
Does it make sense ?
EDIT : I just saw Pickslides' proof. Mine seems a bit redundant now