Binomial Expansion: Find (n+1)Ck in Terms of nCj

[saadsarfraz]

Hi, I've been struggling with this problem for sometime. Let nCk be the kth coefficient in the binomial expansion of (a+b)^n. Find an expression for (n+1)Ck in term of the various nCj. Feel free to treat k=0 and k=n+1 as special cases.

 

[NoMoreExams]

(n+1)Ck = (n+1)!/[(n+1-k)!*k!]

vs.

nCk = n!/[(n-k)!k!]

Now (n+1)! = n!*(n+1) and (n+1-k)! = (n+1-k)(n-k)!

 

[saadsarfraz]

thanks for the reply but i don't think that's the answer. The question I posted is the 1st part. The second part states: Using the expression you found, (which is the question i posted) show that for all n>=0 and for all k, 0=<k<=n, nCk = n!/(k!)(n-k)!

 

[HallsofIvy]

Hi, I've been struggling with this problem for sometime. Let nCk be the kth coefficient in the binomial expansion of (a+b)^n. Find an expression for (n+1)Ck in term of the various nCj. Feel free to treat k=0 and k=n+1 as special cases.

Pascal's triangle: (n+1)Ck= nC(j-1)+ nCj for k not 0.