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Sagot :
An inequality relating 100n and n³ is 100n ≥ n³ for n ≤ 10 and 100n ≤ n³ for n ≥ 10.
What is inequality?
An inequality is comparison of two values, showing if one is less than, greater than, or simply not equal to another value.
Since 100n and n³ for n = 1, 2, 3, . . . 9, 10, 11 are 100, 200, 300, . . . 900, 1000, 1100 and 1, 8, 27, . . . 729, 1000, 1331 respectively.
Therefore, an inequality relating 100n and n³ will be 100n ≥ n³ for n ≤ 10 and 100n ≤ n³ for n ≥ 10.
Induction hypothesis:
Suppose 100n ≤ k³ for some positive integer k ≥ 10.
We need to show that 100( k + 1 ) ≤ ( k + 1 )³ = k³ + 3k² +3k + 1.
Note 100( k + 1 ) = 100k + 100 ≤ k³ + 100
≤ k³ + 3k² (∵ k ≥ 10 )
≤ k³ + 3k² + 3k
≤ k³ + 3k²+3k + 1
So 100( k + 1 ) ≤ ( k + 1 )³, which is true.
Hence by the principle of mathematical induction, 100n ≤ k³ for every integer k ≥ 10.
Know more about inequality here:
https://brainly.com/question/20383699
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