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

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

Promotional Banner

Topper's Solved these Questions

  • PERMUTATIONS AND COMBINATIONS

    MODERN PUBLICATION|Exercise EXERCISE|319 Videos
  • MATHEMATICAL REASONING

    MODERN PUBLICATION|Exercise EXERCISE|169 Videos
  • PROBABILITY

    MODERN PUBLICATION|Exercise EXERCISE|348 Videos

Similar Questions

Explore conceptually related problems

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

Given n^4 <10^n for a fixed integer nge 2 . Prove that (n+1)^4 <10^(n+1) .

Prove that .^(n)C_(0) + (.^(n)C_(1))/(2) + (.^(n)C_(2))/(3) + "……" +(. ^(n)C_(n))/(n+1) = (2^(n+1)-1)/(n+1) .

If a, b, c, d are in G.P, prove that (a^n + b^n), (b^n + c^n), (c^n + d^n) are in G.P.

Prove that (n!)/(r!(n-r)!)+(n!)/((r-1)!(n-r+1)!)= ((n+1)!)/ (r!(n-r+1)!) .

If a,b,c,d are in G.P., prove that (a^(n) + b^(n)), (b^(n) + c^(n)), (c^(n) + d^(n)) are in G.P.

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

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

Prove that : ^(2n)C_n = (2^n [1.3.5. ..........(2n-1)])/(n!) .