Recent content by cupcakes

Exactly. After expanding and canceling out we have: 7x6 - 21x5 + 35x4 - 35x3 + 21x2 - 7x. First I divided by (7x). Then I realized (x-1) is a factor (since x=1 is a zero). After long division I have: (7x)(x-1)(x4 - 2x3 + 3x2 - 2x + 1)

I think I'm very close to solving it. I substituted y= (1-x) in the first equation and expanded. Then I factored that. However I don't know how to factor the last term. x4-2x3+3x2-2x+1 WolframAplha say it can be factored into (x2-x+1)2. I just need to figure out how to factor that into this...

Homework Statement Given: x7 + y7 =1 x + y = 1 Find the integer value(s) of (x-y)2. Homework Equations The Attempt at a Solution I thought of substituting for y and then finding the rational roots but then I realized x and y don't have to be rational numbers for (x-y)^2 to be an...

Homework Statement To prove the inequality (attached) Homework Equations The Attempt at a Solution I tried factoring out a 2 from each of the even terms in the denominator. This allowed me to cancel out all the terms (odd) on the numerator up to 1005. Leaves me with...

I've solved it. :smile:

I forgot to mention that the problem states that it must be proven without using induction. :( Thanks Sammy :smile: Does anyone have any other idea or hint that does not involve induction?

Homework Statement \frac{1^2*3^2*5^2...(2n-1)^2}{2^2*4^2*6^2...(2n)^2}<\frac{1}{2n+1} Edit: Must be proven without using induction. Homework Equations The Attempt at a Solution I understand the LHS is the same thing as \frac{(2n-1)!}{(2n)!} And (2n)! = k!2^k & (2n-1)! =...

that seems right to me. However I'd keep sin(π/3) as \frac{\sqrt{3}}{2}. Edit: I didn't see rock.freak667's reply.

Thank you, I finally understand what it means. It may be easy for you but I've never learned permutations or anything like this. This problem is not course material, just something I came across. Alright that makes sense. Since we can prove that there will always be at least one even factor...

From what Ray said, I gathered that we substitute arrangements of the 2011 numbers. However, if I understand you correctly, you're saying that we substitute the integers themselves. I'm confused... Edit An attempt at the problem: ODD + ODD = EVEN ODD + EVEN = ODD...

So let's say instead of 2 to 2012 its from 7 to 12. would one possible arrangement be {10,12,9,7,8,11} → 101297811 ? if that's so I'm guessing there are much more possible arrangements of 2 to 2012 than 2012. Yet the product only has up to a2012. So is this some sort of probability question...

How did I not notice that?! :smile: That works perfectly Mark. Thanks to everyone for their help.

Homework Statement Please see attached image. Homework Equations The Attempt at a Solution I don't understand the problem at all. Can someone explain to me what the problem is stating, more precisely what an arrangement is? An example would be nice. Thanks.

I may be wrong but in mathematics, n is used to denote an integer by convention.

Maybe it's because I forgot to mention that A & B may contain roots. I think it's possible because it is an assigned question. But I'm stuck...