Hey puresoul.
Hint: (r+1)*r! = (r+1)!
Can someone please help me with this question
Use the inductive property:
nC(r+1) = (n-r)/(r+1) * nCr
to prove the Pascal triangle property that nCr + nC(r+1) = (n+1)C(r+1)
This is how I started..
nCr + (n-r)/(r+1) * nCr = (n+1-r)/(r+1) * nCr -->
nCr (1+ (n-r)/(r+1)) = nCr (n+1-r)/(r+1)
I don't know what to do next..
I'm going to attempt it straight from the Pascal relationship:
nCr + nC(r+1)
= n!/[r!*(n-r)!] + n!/[(r+1)!*(n-r-1)!]
= n!/[r!*(n-r-1)!] * [1/(n-r) + 1/(r+1)]
= n!/[r!*(n-r-1)!] * [r+1 + n - r]/[(n-r)(r+1)]
= n!/[r!*(n-r-1)!] * [n+1]/[(n-r)(r+1)]
= (n+1)!/[(r+1)!*(n-r)!]
= (n+1)C(r+1)