hi every1..
can anyone prove that:
(Coff. X^(r+1) - Coff. X^r) in the expansion (1 + X)^(n+1)
is equal to
(Coff. X^(r+1) - Coff. X^(r-1)) in the expansion (1 + X)^n
i tried to solve it but i go around in circles, can anyone help? (10 pts for the right answer)
thanks..
2006-12-24 21:29:37 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
but why does (n+1 C r+1) = n C r + (n C r+1)??
2006-12-24 22:31:52 · update #1
its easy...
(n+1 C r+1) = n C r + (n C r+1)
LHS = nCr + (nCr+1) -(n+1Cr)
Also (n+1Cr) = nCr+ (nCr-1) so nCr-(n+1Cr) = -(nCr-1)
So LHS = (nCr+1) - (nCr-1)
=RHS
Merry Christmas
2006-12-24 21:55:21 · answer #1 · answered by surya o 2 · 0⤊ 0⤋
The first line gives you (n+1)! / (n-r)!(r+1)! - (n+1)! / (n+1-r)! r!, and the second line n!/ (n-r-1)!(r+1)! - n! / (n+1-r)!(r-1)!. All you have to do
is reduce to the same denominator.
You can factor out n! on the top and (n-r-1)!(r-1)! on the bottom.
That leaves you with (n+1)/(n-r)(r+1)r - (n+1)/(n+1-r) (n-r) r and
1 / (r+1)r - 1 / (n+1-r)(n-r). You are reduced to something you can do.
2006-12-25 06:07:21 · answer #2 · answered by gianlino 7 · 0⤊ 0⤋
why are you concerned with that crap on christmas day?
2006-12-25 05:36:48 · answer #3 · answered by Mr. Nice Guy 1 · 0⤊ 2⤋