The question is to write the following as a general result about
binomial coefficients of which this is a special case,
2C2 + 3C2 + 4C2 +5C2 +6C2 = 7C3.
I can write a general result but how to go about proving it?
The above can be written as rCr + (r+1)Cr + (r+2)Cr + (r+3)Cr
+...+ nCr = (n+1)C(r+1).
To prove it I should start from the left as it is the messier side
but I have no idea where to begin. From the right I can expand and
get
nCr + nC(r+1) by using (n+1)C(r+1)= nCr + nC(r+1)
I can continue this for nC(r+1)= (n-1)C(r+1) + (n-1)Cr.
Then (n-1)C(r+1)= (n-2)C(r+1) + (n-2)Cr
I can see that the right hand will expand out to equal the left but
it is long and tedious.
How can the L.H.S be simplified to show it equals the right?
Thanks.