Find the sum of `1/(1!(n-1)!)+1/(3!(n-3)!)+1/(5!(n-5)!)+ ...,`

Explanation

`because 1 ! = 1 `
` therefore ` The given series can be written as
`(1)/(n!)(1)/((n-1)!1!) + (1!)/((n-3)!3!) + (1)/((n-5)!5!) + ...` …(i)
` because ` Sum of values of each terms in factorial are equal
.` (n-1) + 1 = (n-3) + 3= (n-5) + 5 = … = n `
From Eq. (i)
`(1)/(n!)[(1)/((n-1)!1!) + (1!)/((n-3)!3!) + (1)/((n-5)!5!)+ ...]`
` = (1)/(n!) ( ""^(n)C_(1) + ""^(n)C_(3) + ""^(n)C_(3) + ""^(n)C_(5) + ...) = (2^(n-1))/(n!)` .