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17 August, 15:00

Prove by induction that n! ≤ n^n

for all n ∈ N.

+2
Answers (2)
  1. 17 August, 15:09
    0
    n! ≤ n^n

    Step-by-step explanation:

    n! ≤ n^n

    Proof

    let n=1

    1!=1=1^1=1

    hence 1=1

    when n=2

    2!=1x2=2 and 2^2 = 2x2=4

    hence 2≤4

    when n=n+1, (n+1) !=n! (n+1) = (n+1) ^ (n+1) = (n+1) ^n x (n+1)

    i. e. n! (n+1) = (n+1) ^nXn+1

    Divide both sides by n+1

    n! = (n+1) ^n

    hence n! ≤ n^n
  2. 17 August, 16:07
    0
    the equation given satisfies the given condition of n!<=n^n

    Step-by-step explanation:

    taking n=4 and n=2 and n=1

    4! < = 4^4

    4*3*2*1 < = 256

    24 <256

    2! < = 2^2

    2*1 < = 4

    2 < 4

    1! < = 1^1

    1*1 < = 1

    1=1

    hence proved
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