Recent content by steveT

But I don't have the LHS = RHS anywhere. For a proof like this I need to get the original LHS to look like the original RHS, then I need to get the original RHS to look like the original LHS. I have done neither.

On the LHS I'm stuck at A^c n B^c n C, I'm trying to get it to equal the original RHS. As far as everything I did on the RHS, I was trying to get it to equal the LHS

(A U B)^c - C^c = A^c - (B U C^c) A^c n B^c n C = (A^c n B^c) n (A^c n C) A^c n B^c n C = (A U B)^c n (A^c n C) A^c n B^c n C = (A U B)^c n (A^c - C^c) I'm getting close, is this correct so far. Not sure where to go now?

How about this (A U B)^c - C^c = A^c n B^c n C Still not sure how to make that look like the RHS of the equation

OK, here is what I got so far, still trying to prove the RHS (A U B)^c - C^c A^c n B^c - C^c Since A^c n B^c = A^c - B I get A^c - B - C^c Now I'm stuck again

Wow, I just wasted a lot of time, I wrote it down wrong. It should have been this (A U B)^c - C^c = A^c - (B U C^c) Ok, now that that is settled, I'm trying to prove the RHS first given the LHS. Like I stated before, I know that (A U B)^c = A^c and B^c So I get A^c and B^c -...

What part is generally false? Maybe this is more clear. The complement of ( A U B) - The complement of C = The complement of A - The complement of (B U C) I know that the complement of ( A U B ) = the complement of A and the complement of B

Homework Statement I need to prove that: (A U B)^c - C^c = A^c - (B U C)^c Homework Equations The Attempt at a Solution I know that (A U B)^c = A^c and B^c My problem is I'm not sure how to rearrange or distribute the minus sign to make it equal to the other side...

Thanks everyone for your help. I UNDERSTAND

So when k is even you get plus and when its odd you get minus which accounts for the alternating sign.

x=1 and a=0 ?

This is the expression I'm using (x+a)^n=∑_(k=0)^n▒〖(n¦k) x^k a^(n-k) 〗

Well I've been staring at this thing for the past hour and I'm not coming up with anything. Am I to be looking at the (x+y)^n side of the binomial theorem or the side with the summation

Homework Statement (n¦0)-(n¦1)+(n¦2)-. . . ± (n¦n)=0 that reads n choose zero and so on Homework Equations Prove this using the binomial theorem The Attempt at a Solution I really have no idea where to start. Any help would be greatly appreciated thanks

Can anyone help me out with this one? I need to find an explicit formula for: n 2 ∑ i i=1 I was already asked to find the explicit polynomial formula for the above equation which is n·(n + 1)·(2·n + 1) _________________ 6 I'm not sure on what the difference...