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$\frac{1}{1! (n - 1)!} + \frac{1}{3! (n - 3)!} + \frac{1}{5 (n - 5)!} + \dots$ is equal to

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2nn!

2n−1n!

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detailed solution

Correct option is C

Given that, 11!(n−1)!+13!(n−3)!+15!(n−5)!+…=1n!n!1!(n−1)!+n!3!(n−3)!+n!5!(n−5)!+…=1n! nC1+nC3+nC5+…=2n−1n!

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11!(n−1)!+13!(n−3)!+15(n−5)!+… is equal to

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$\frac{1}{1! (n - 1)!} + \frac{1}{3! (n - 3)!} + \frac{1}{5 (n - 5)!} + \dots$ is equal to