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Question: \[\sum_{r = 0}^{m}{n + rC_{n} =}\]...

∑r=0mn+rCn=\sum_{r = 0}^{m}{n + rC_{n} =}

A

n+m+1Cn+1n + m + 1C_{n + 1}

B

n+m+2Cnn + m + 2C_{n}

C

n+m+3Cn−1n + m + 3C_{n - 1}

D

None of these

Answer

n+m+1Cn+1n + m + 1C_{n + 1}

Explanation

Solution

Since nCr=n⥂Cn−rnC_{r} =^{n} ⥂ C_{n - r} and nCr−1+n⥂Cr=n+1⥂CrnC_{r - 1} +^{n} ⥂ C_{r} =^{n + 1} ⥂ C_{r} we have

∑r=0mn+rCn=∑r=0mn+rCr=n⥂C0+n+1⥂C1+n+2⥂C2+......+n+m⥂Cm\sum_{r = 0}^{m}{n + rC_{n}} = \sum_{r = 0}^{m}{n + rC_{r}} =^{n} ⥂ C_{0} +^{n + 1} ⥂ C_{1} +^{n + 2} ⥂ C_{2} + ...... +^{n + m} ⥂ C_{m}

=[1+(n+1)]+n+2⥂C2+n+3⥂C3+........+n+m⥂Cm= \lbrack 1 + (n + 1)\rbrack +^{n + 2} ⥂ C_{2} +^{n + 3} ⥂ C_{3} + ........ +^{n + m} ⥂ C_{m}

=n+m+1⥂Cn+1=^{n + m + 1} ⥂ C_{n + 1}, [∵ 6munCr=n⥂Cn−r]\lbrack\because\mspace{6mu}^{n}C_{r} =^{n} ⥂ C_{n - r}\rbrack.