Solving the Inequality \(n^2 < 2^n\) for Natural Numbers

The question asks for which condition on the natural number \(n\) the inequality \(n^2 < 2^n\) is true for all natural numbers satisfying that condition.

Let's test the inequality for the first few natural numbers to see when it holds true:

• For \(n=1\): \(1^2 = 1\), \(2^1 = 2\). Is \(1 < 2\)? Yes.
• For \(n=2\): \(2^2 = 4\), \(2^2 = 4\). Is \(4 < 4\)? No.
• For \(n=3\): \(3^2 = 9\), \(2^3 = 8\). Is \(9 < 8\)? No.
• For \(n=4\): \(4^2 = 16\), \(2^4 = 16\). Is \(16 < 16\)? No.
• For \(n=5\): \(5^2 = 25\), \(2^5 = 32\). Is \(25 < 32\)? Yes.
• For \(n=6\): \(6^2 = 36\), \(2^6 = 64\). Is \(36 < 64\)? Yes.

We can summarize these results in a table:

From the table, we observe that the inequality \(n^2 < 2^n\) is true for \(n=1\) and for \(n=5, 6, \dots\). It is false for \(n=2, 3, 4\). We can see a pattern suggesting that for \(n \ge 5\), the inequality \(n^2 < 2^n\) holds true.

Now let's look at the given options:

• Option 1: \(n \ge 5\). This condition includes natural numbers 5, 6, 7, and so on. Our testing suggests that for all these numbers, \(n^2 < 2^n\) is true.
• Option 2: \(n < 5\). This condition includes natural numbers 1, 2, 3, and 4. Our testing shows that the inequality is false for \(n=2, 3, 4\). Thus, it is not true for all natural numbers satisfying this condition.
• Option 3: \(n \le 3\). This condition includes natural numbers 1, 2, and 3. Our testing shows that the inequality is false for \(n=2, 3\). Thus, it is not true for all natural numbers satisfying this condition.

The condition under which the inequality \(n^2 < 2^n\) is true for all natural numbers meeting that condition is \(n \ge 5\).