Question

What value for the positive integer k for which $\displaystyle \sf \frac{(101)^{k/2}}{k!}$ is a maximum.

Answer 1

k=10, will give max. value...............................

Answer 2

Answer:

f(k)=(101)k2k!Let maximum value occurs for k=n, thenf(n)>f(n+1)⇒(101)n2k!>(101)n+12(n+1)!⇒(101)n2n!>(101)n2(101)12(n+1) n!⇒1>(101)12(n+1) ⇒n+1>(101)12⇒n>(101)12−1Now (101)12 is slightly greater than 10 [As (100)12=10] which means(101)12−1 is slightly greater than 9, hence next integer after 9 isn=10, Hence=k=10.