Home
Class 11
MATHS
Prove that (n !)^2 < n^n n! < (2n)!, for...

Prove that `(n !)^2` < `n^n n!` < `(2n)!`, for all positive integers n.

Promotional Banner

Similar Questions

Explore conceptually related problems

Prove that (2n)^2 where n in N can be expressed as the sum of n terms of a series of integers in AP

Prove that (n!) (n + 2) = [n! + (n + 1)!]

Prove that ((2n)!)/(2^(2n)(n!)^(2))<=(1)/(sqrt(3n+1)) for all n in N

Prove that : (2n) ! = 2^n (n!)[1.3.5.... (2n-1)] for all natural numbers n.

Prove that ((2n)!) / (n!) = 2^n(2n - 1) (2n - 3) ... 5.3.1.

Prove that n! ( n +2) = n ! + ( n + 1) !

Prove that: n !(n+2)=n !+(n+1)!

Prove that n! (n+2) = n! +(n+1)! .

Prove that n! (n+2) = n! +(n+1)! .

Prove that [(n+1)//2]^n >(n !)dot