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B
2nCn−1
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C
2nCn+1
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D
2n+1Cn−1
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Solution
The correct option is A2n+1Cn The given series can be written as nC0+n+1C1+n+2C2+.....+2nCn={n+1C0+n+1C1}+n+2C2+.....+2nCn={n+2C1+n+2C2}+.....+2nCn={n+3C2+n+3C3}+n+4C4+....+2nCn=n+4C3+n+4C4+....+2nCn Proceeding the same way finally we get the sum as 2n+1Cn