Respuesta :

moizmubeen

Answer:

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

funkeyusuff52

Answer:

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

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