prove nC(r-1) + (n+1)C(r-1) + (n+2)C(r-1)+ ... +(n+k)C(r-1) = (n+k+1)Cr
To prove the equation (nC(r-1)) + ((n+1)C(r-1)) + ((n+2)C(r-1))+ ... +((n+k)C(r-1)) = (n+k+1)Cr, we can use the concept of combinatorial identities and the Pascal's identity.
Pascal's Identity states that for any positive integer n and any integer r between 0 and n, the following equation holds:
nCr = (n-1)C(r-1) + (n-1)Cr
Using Pascal's Identity, we can rewrite the left side of our equation as follows:
(nC(r-1)) + ((n+1)C(r-1)) + ((n+2)C(r-1))+ ... +((n+k)C(r-1))
= [(n+1-1)C((r-1)-1)] + [(n+2-1)C(r-1)] + [(n+3-1)C(r-1)] + ... + [(n+k-1)C(r-1)] + [(n+k)C(r-1)]
= (nC(r-2)) + ((n+1)C(r-1)) + ((n+2)C(r-1))+ ... +((n+k-1)C(r-1)) + ((n+k)C(r-1))
We observe that the terms in the equation are in the form (n+p)C(r-1), where p ranges from 0 to k. Notice that (n+k+1)-(n-1) = (n+p) for each term.
So we can rewrite the equation using the new variable q = p + n - 1:
= (nC(r-2)) + ((n+1)C(r-1)) + ((n+2)C(r-1))+ ... +((n+k-1)C(r-1)) + ((n+k)C(r-1))
= ((n+1-1)C((r-1)-1)) + ((n+2-1)C((r-1)-1)) + ((n+3-1)C((r-1)-1)) + ... + ((n+k-1)C((r-1)-1)) + ((n+k)C((r-1)-1))
= ((n+1)-1)C(((r-1)-1)+n-1)) + ((n+2)-1)C(((r-1)-1)+n-1)) + ((n+3)-1)C(((r-1)-1)+n-1)) + ... + ((n+k)-1)C(((r-1)-1)+n-1)) + ((n+k)-1)C(((r-1)-1)+n-1))
= (n+1)Cr + (n+2)Cr + (n+3)Cr + ... + (n+k)Cr
Notice that the expression we have now matches the right side of the equation: (n+k+1)Cr.
Therefore, we have successfully proven that:
(nC(r-1)) + ((n+1)C(r-1)) + ((n+2)C(r-1))+ ... +((n+k)C(r-1)) = (n+k+1)Cr.