First slide

Combinations

Question

$\sum_{r = 0}^{m} {}^{n + r}C_{n} is equal to:$

Difficult

A

${}^{n + m + 1}C_{n + 1}$
B

${}^{n + m + 2}C_{n}$
C

${}^{n + m + 3}C_{n - 1}$
D

None of these

Solution

$\begin{matrix} \sum_{r = 0}^{m} {}^{n + r}C_{n} = \sum_{r = 0}^{m} {}^{n + r}C_{r} (∵^{n + r} C_{n} =^{n + r} C_{n + r - n}) \\ =^{n} C_{0} +^{n + 1} C_{1} +^{n + 2} C_{2} +^{n + 3} C_{3} + \dots \dots +^{n + m} C_{m} \\ {Using,}^{n} C_{0} = 1 =^{n + 1} C_{1} \\ = (^{n + 1} C_{0} +^{n + 1} C_{1}) +^{n + 2} C_{2} +^{n + 3} C_{3} + \dots . +^{n + m} C_{m} \\ Using this again and again, we are left \\ = {}^{n + m + 1}C_{m} = {}^{n + m + 1}C_{n + 1} \end{matrix}$

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