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  4. For n ∈ N, 10n-2 ≥ 81 n, is

Q. For $n \in N, 10^{n-2} \geq 81\, n$, is

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Solution:

Let $P(n): 10^{n-2} \geq 81\, n$
For $n=4,\,\,\,10^{2} ⪈ 81 \times 4$
For $n=5,\,\,\,10^{3} \ge 81 \times 5$
Hence, by mathematical induction for $n \geq 5$,the proposition is true.

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