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\(\displaystyle\sum_{r=1}^{m}\)n+rC

Write ∑n+rCr r∈(0,m) in the simplified form.

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Given,

n+rCr r∈(0,m)

We know : 

nCr + nCr-1 = n+1Cr ....(1)

\(\sum_{r=0}^{m} \) n+rC= 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+rC= n+2C1 + n+2C2 + n+3C3 + . . . . . . + n+mCn+1

 \(\sum_{r=0}^{m} \)n+rC= n+3C2 + n+3C3 + . . . . . . + n+mCn+1

Proceeding in the same way :

 \(\sum_{r=0}^{m} \)n+rCn+mCm-1 + n+mCm = n+m+1Cn+1

 \(\sum_{r=0}^{m} \)n+rCn+m+1Cn+1

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