Recent content by dystplan

wait; 2^100/100!

wait; 2^100/100!

Unless... x/y = 0 or 1 in the binomial theorem? That would make it easy. Makes it = 1/100! ?

well; thanks. But I'm still just totally lost. (1+1)^100 is a massive equation when expanded.

C(100,n) being = 100!/n!(100-n)! ? hmmm, unfortunately I don't see where that's going =/ Man this one is throwing me for a loop, only question I haven't managed and due tomorrow.

Homework Statement \sum_{n=0}^{100} 1/n!(100-n)! The Attempt at a Solution Other then obvious attempts to make sense of the equation's incremental and decrements divisor, I can't figure out where to start with this question. Some assistance would be greatly appreciated.

The only way I know to discover B(n+1) is to have a starting place and identify a pattern to progression. I understand that the question isn't asking me to count them, but how can I calculate the recursion without a base set that indicates a recursive relationship?

I get B(1) = 1 B(2) = 3 B(3) = 12 B(4) = 36 B(5) = 171 B(6) = 648 I still don't see any way that's recursive =[ This problem has been driving me nuts all day. Where X represents any chip but B: 5 stack: XXXXB x 81 possibilities BXBXB x 9 possibilities BXXXB x 27 XBXXB x 27...

You mean something like: X1 = 4 X2 = 12 + 4 = 16; x3 = 48 + 4*4 = 64 x4 = 192 + 4*4*4 = 256 X1 = 4 Xn = X(n-1)*4 + 4^(n-1) The problem is that I don't see any correlation with the rate at which exclusions occur due to consecutive blue poker chips. 0, 1, 7, 34, 145 (unless I'm doing...

Thanks for the help Dick.

Thanks; I've fixed that problem; durh. Still have no idea how to get from 1/(4k^2+8k+3)−1) (or 1/((2k+3)(2k+1)) ) to (k+1)/(2*(k+1)+1)

Homework Statement Construct a recursive definition for the following: The number of ways to stack n red, white, green, and blue poker chips so that there are no two consecutive blue poker chips. The attempt at a solution The number of possible combinations is 4^n minus the number of...

Homework Statement Prove by mathematical induction: \sum^n_{i=1} 1/((2i-1)(2i+1)) = n/(2n+1) The attempt at a solution \sum^k+1_{i=1} = 1/((2k-1)(2k+1)) + 1/((2(k+1)-1)(2(k+1)+1)) = k/(2k+1) + 1/(4(k+1)^2 + 2(k+1) - 2(k+1) -1) = k/(2k+1) + 1/(4(k+1)^2 -1) = k/(2k+1) + 1/(4(k^2 + 2k + 1)) =...