C7 14+∑i=13C6 17-i=

14C7+∑i=1317−iC6=14C7+14C6+15C6+16C6=15C7+15C6+16C6 ∵Cr n+Cr-1 n=Cr n+1 =16C7+16C6 =17C7

C7 14+∑i=13C6 17-i=

14C7+∑i=1317−iC6=14C7+14C6+15C6+16C6=15C7+15C6+16C6 ∵Cr n+Cr-1 n=Cr n+1 =16C7+16C6 =17C7

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${}^{14}C_{7} + \sum_{i = 1}^{3} {}^{17 - i}C_{6} =$

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${}^{16}C_{8}$

${}^{16}C_{7}$

${}^{17}C_{7}$

${}^{17}C_{8}$

answer is B.

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Detailed Solution

$\begin{array}{l} ^{14} C_{7} + {\sum_{i = 1}^{3}}^{17 - i} C_{6} =^{14} C_{7} +^{14} C_{6} +^{15} C_{6} +^{16} C_{6} =^{15} C_{7} +^{15} C_{6} +^{16} C_{6} (∵ {}^{n}C_{r} + {}^{n}C_{r - 1} = {}^{n + 1}C_{r}) \\ =^{16} C_{7} +^{16} C_{6} \\ =^{17} C_{7} \end{array}$