For every natural number $n \geq 2,$
Statement 1: $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n}$
Statement 2: $\sqrt{n (n + 1)} < n + 1$

Statement 1 is false, Statement 2 is true.

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Statement 1 is true, Statement 2 is true. Statement 2 is a correct explanation for Statement 1.

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Statement 1 is true, Statement 2 is true. Statement 2 is not a correct explanation for Statement 1.

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Statement 1 is true, Statement 2 is false.

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Solution

The correct option is C Statement 1 is true, Statement 2 is true. Statement 2 is not a correct explanation for Statement 1.
$Let P (n) be the statement, P (n) : \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n} Consider P (n) for n = 2 \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} = 1 + \frac{1}{\sqrt{2}} > \sqrt{2} Assume P (k) is true, i . e ., \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{k}} > \sqrt{k} . . . (1) Consider \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}} > \sqrt{k} + \frac{1}{\sqrt{k + 1}} (Using (1)] = \frac{\sqrt{k (k + 1)} + 1}{\sqrt{k + 1}} > \frac{k + 1}{\sqrt{k + 1}} (Since \sqrt{k (k + 1)} > k, \forall k \geq 1] = \sqrt{k + 1} i . e ., \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}} > \sqrt{k + 1} Thus, P (k + 1) is true whenever P (k) is true . Hence, by principle of mathematical induction, P (n) is true for all n \geq 2 .$
$Now consider the second statement . P (n) : \sqrt{n (n + 1)} < n + 1 For n = 2, \sqrt{2 (2 + 1)} < 2 + 1 i . e ., \sqrt{3} < 3 which is true . Assume P (k) is true . Then, \sqrt{k (k + 1)} < k + 1 Now, consider \sqrt{(k + 1) (k + 1 + 1)} = \sqrt{(k + 1) (k + 2)} < \sqrt{(k + 2) (k + 2)} (Since k + 1 < k + 2] = k + 2 i . e ., \sqrt{(k + 1) (k + 2)} < k + 2 Therefore, by principle of mathematical induction, P (n) is true for all n \geq 2 . So, statement 1 and statement 2 both are true . But we don' t need to use statement 2 in order to prove / derive statement 1 . Therefore, Statement 2 is not a correct explanation for Statement 1 .$

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