nC0+nC2+nC4+…⋯+nC2[n/2], where [ ] denotes greatest integer
A
22n−1
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B
22n−1−1
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C
2n−1
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D
2n−1−1
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Solution
The correct option is C2n−1 The real sol. is taking a binomial = (1+x)n=nC0+nC1x+nC2x2+...... and now taking (1−x)n=nC0−nC1x+nC2x2−...... Now adding both and putting x=1, we get 2n=2(nC0+nC2+nC4+....) => 2n−1 = required expression