G GaganDeep fffff Joined Oct 12, 2005 Messages 109 Gender Male HSC 2006 Sep 23, 2006 #1 Show that (nCr)nCn-1)=(n-r+1) Hence find the sum of nC1:nC0 + 2(nC2):nC1 +3(nC3):nC2 + ...+n (nCn):nCn-1 I can do first part but can't do the sum
Show that (nCr)nCn-1)=(n-r+1) Hence find the sum of nC1:nC0 + 2(nC2):nC1 +3(nC3):nC2 + ...+n (nCn):nCn-1 I can do first part but can't do the sum
R Riviet . Joined Oct 11, 2005 Messages 5,593 Gender Undisclosed HSC N/A Sep 24, 2006 #2 I think the first part was supposed to read: Show that r(nCr)/{nC(r-1)} = n-r+1 Sum = (n-1+1) + (n-2+1) + ... + (n-(n-1)+1) + (n-n+1) [using first part] = n + (n-1) + (n-2) + ... + 2 + 1 This is a sum of an AP with n terms, d=-1, a=n .'. Sum = n/2.(2a + (n-1)d) =(n/2).(2n + (n-1)(-1)) =(n/2).(2n-n+1) =(n/2).(n+1) Last edited: Sep 24, 2006
I think the first part was supposed to read: Show that r(nCr)/{nC(r-1)} = n-r+1 Sum = (n-1+1) + (n-2+1) + ... + (n-(n-1)+1) + (n-n+1) [using first part] = n + (n-1) + (n-2) + ... + 2 + 1 This is a sum of an AP with n terms, d=-1, a=n .'. Sum = n/2.(2a + (n-1)d) =(n/2).(2n + (n-1)(-1)) =(n/2).(2n-n+1) =(n/2).(n+1)
nCn-1)=(n-r+1)