pls i need someone to help prove these.
\(proof\) that:
1) nCr=n−1Cr+n−1Cr−1^{n}C_{r} = ^{n-1}C_{r} +^{n-1}C_{r-1}nCr=n−1Cr+n−1Cr−1.
2) ∑r=0kmCr+nCk−r=m+nCk\sum {^{k}_{r=0}} ^{m}C_{r} + ^{n}C_{k-r} =^{m+n}C_{k}∑r=0kmCr+nCk−r=m+nCk.
i'll be grateful if anyone can help with these!!!
Note by Samuel Ayinde 6 years, 2 months ago
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1) n−1Cr+n−1Cr−1^{n-1}C_{r}+^{n-1}C_{r-1}n−1Cr+n−1Cr−1
=(n−1)!r!(n−r−1)!+(n−1)!(r−1)!(n−r)!=\frac{(n-1)!}{r!(n-r-1)!}+\frac{(n-1)!}{(r-1)!(n-r)!}=r!(n−r−1)!(n−1)!+(r−1)!(n−r)!(n−1)!
=(n−1)!(r−1)!(n−r−1)!(1r+1n−r)=\frac{(n-1)!}{(r-1)!(n-r-1)!}\left(\frac{1}{r}+\frac{1}{n-r}\right)=(r−1)!(n−r−1)!(n−1)!(r1+n−r1)
=(n−1)!(r−1)!(n−r−1)!(nr(n−r))=\frac{(n-1)!}{(r-1)!(n-r-1)!}\left(\frac{n}{r(n-r)}\right)=(r−1)!(n−r−1)!(n−1)!(r(n−r)n)
=n!r!(n−r)!=nCr=\frac{n!}{r!(n-r)!}=^{n}C_{r}=r!(n−r)!n!=nCr
2) Make use of this: ∑r=0knCk−r=∑r=0knCr\displaystyle\sum{_{r=0}^{k}}^{n}C_{k-r}=\displaystyle\sum{_{r=0}^{k}}^{n}C_{r}∑r=0knCk−r=∑r=0knCr
I can't seem to find a solution. Do reply if you manage to prove it!
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Comments
1) n−1Cr+n−1Cr−1
=r!(n−r−1)!(n−1)!+(r−1)!(n−r)!(n−1)!
=(r−1)!(n−r−1)!(n−1)!(r1+n−r1)
=(r−1)!(n−r−1)!(n−1)!(r(n−r)n)
=r!(n−r)!n!=nCr
2) Make use of this: ∑r=0knCk−r=∑r=0knCr
I can't seem to find a solution. Do reply if you manage to prove it!