Given,
∑n+rCr r∈(0,m)
We know :
nCr + nCr-1 = n+1Cr ....(1)
\(\sum_{r=0}^{m} \) n+rCr = nC0 + n+1C1 + n+2C2 + n+3C3 + . . . . . . + n+mCn+1
\(\sum_{r=0}^{m} \)n+rCr = n+1C0 + n+1C1 + n+2C2 + n+3C3 + . . . . . . + n+mCn+1
⇒ (nC0 = n+1C0)
Using equation (1),
\(\sum_{r=0}^{m} \)n+rCr = n+2C1 + n+2C2 + n+3C3 + . . . . . . + n+mCn+1
\(\sum_{r=0}^{m} \)n+rCr = n+3C2 + n+3C3 + . . . . . . + n+mCn+1
Proceeding in the same way :
\(\sum_{r=0}^{m} \)n+rCr = n+mCm-1 + n+mCm = n+m+1Cn+1
\(\sum_{r=0}^{m} \)n+rCr = n+m+1Cn+1