Thread: binomial expansion

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..

Hey puresoul.

Hint: (r+1)*r! = (r+1)!

Originally Posted by chiro

Hey puresoul.

Hint: (r+1)*r! = (r+1)!

Thnks for your hint..

I took another approach..

n!/(r!(n-r)!) + n!/ ( (r+1)!(n-(r+1))!

Then i tried to multiply the first fraction by (r+1) so i got in the denominator (r+1)!, but i didn't know how to simplify the top part and to get (n+1)!

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)

Originally Posted by chiro

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)

THNKS A LOT ^_____^
I got it !!