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B
(n+k+1)Ck
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C
(n+k)Ck−1
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D
(n+k−1)Ck
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Solution
The correct option is D(n+k−1)Ck k−1Ck+kCk−1–––––––––––––––+....+n+k−2Ck−1 =k+1Ck+k+1Ck−1––––––––––––––––––+....+n+k−2Ck−1 =k+2Ck+...............+n+k−2Ck−1=n+k−1Ck