f-n-1-x-f-n-x-n-1-n-f-0-x-1-f-5-x-

Question Number 377 by 123456 last updated on 25/Jan/15

f_{n + 1} (x) = \frac{f_{n} (x)}{n + 1} + n

f_{0} (x) = 1

f_{5} (x) = ?

Answered by prakash jain last updated on 25/Dec/14

f_{0} (x) = 1

f_{1} (x) = \frac{1}{1} + 0 = 1

f_{2} (x) = \frac{1}{2} + 1 = \frac{3}{2}

f_{3} (x) = \frac{\frac{3}{2}}{2 + 1} + 2 = \frac{1}{2} + 2 = \frac{5}{2}

f_{4} (x) = \frac{\frac{3}{2}}{3 + 1} + 3 = \frac{3}{8} + 3 = \frac{27}{8}

f_{5} (x) = \frac{\frac{27}{8}}{4 + 1} + 4 = \frac{27}{40} + 4 = \frac{27 + 160}{40} = \frac{187}{40}