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B
2n+1
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C
3⋅2n
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D
2n−1
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Solution
The correct option is C2n−1 Simplifying, we get (n+1C1+n+1C2+...n+1Cn)−(nC1+nC2+...nCn) =(n+1C1+n+1C2+...n+1Cn+n+1Cn+1−n+1Cn+1)−2n =2n+1−2n−n+1Cn+1 =2n(2−1)−1 =2n−1