Question

Consider the function f defined on the set of all non-negative interger such that `f(0) = 1, f(1) =0` and `f(n) + f(n-1) = nf(n-1)+(n-1) f(n-2)` for `n ge 2`, then f(5) is equal to
A. 40
B. 44
C. 45
D. 60

Answer 1

Correct Answer - B
We have,
`f(n) + f(n-1) = nf(n-1) + n(n-1) f(n-2)`
`rArr f(n) - nf(n-1) = - { f(n-1) - (n-1) f(m-2)}`
`=(-1)^(2) {f(n-2) = - {f(n-1) -(n-1) f(n-2)}`
` = (-1)^(n-1) {(f(1) - f(0)}`
`rArr f(n) - n f(n-1) = - (1)^(n)`
`rArr (f(n))/(n!) = (f(n-1))/((n-1)!) +((-1)^(n-1))/((n-1)!)`
`rArr (f(n-2))/((n-2)!) = (f(n-3))/((n-3)!)+((-1)^(n-2))/((n-2)!)`
`(f(2))/(2!) = (f(1))/(1!) +((-1)^(2))/(2!)`
`(f(1))/(1!) = (f(0))/(0!) +((-1)^(1))/(1!)`
Adding all these equalities, we obtain
`(f(n))/(n!) = ((-1)^(n))/(n!)) +((-1)^(n-1))/((n-1)!) +((-1)^(n-2))/((n-2)!) +......+((-1)^(2))/(2!)+((-1)^(1))/(1!)`
` f(n) = n! {1-(1)/(1!)+(1)/(2!) -(1)/(3!)+....+((-1)^(n))/(n!)}`
`therefore f(5) = 5! (1-(1)/(1!)+(1)/(2!) -(1)/(3!) + (1)/(4!)-(1)/(5!))= 44`