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B
3n−1
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C
3n+1+1
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D
None of these
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Solution
The correct option is A3n−1 Expanding, we get nC1(1C0+1C1)+nC2(2C0+2C1+2C2)+....nCn(nC1+nC2+...nCn) =nC121+nC222+...nCn2n =nC020+nC121+nC222+...nCn2n−[nC020] =(1+2)n−1 =3n−1