InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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 toA. 40B. 44C. 45D. 60 |
|
Answer» 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` |
|