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Mathematics
Norah
17 August, 15:00
Prove by induction that n! ≤ n^n
for all n ∈ N.
+2
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2
)
Morgan Blackburn
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
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Saunders
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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