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...
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