For every natural number n-2-Statement 1- 11-12-1n-nStatement 2- nn-1-n-1

Let nbsp;Pn nbsp;be nbsp;the nbsp;statement,Pn: nbsp;11+12+...+1n gt;nConsider nbsp;Pn nbsp;for nbsp;n=211+12=1+12 gt;2Assume nbsp;Pk nb ...

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Standard XII

Mathematics

Statement

For every nat...

Question

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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Similar questions

Statement I

For every natural number

n \geq 2

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} > \sqrt{n}

Statement II

For every natural number

n \geq 2

\sqrt{n (n + 1)} < n + 1

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

Q. Statement-l: For every natural number

n \geq 2, \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots \dots + \frac{1}{\sqrt{n}} > \sqrt{n}

.
Statement-2: For every natural number

n \geq 2, \sqrt{n (n + 1)} < n + 1

Q. 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

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For every natural number n≥2, Statement 1: 1√1+1√2+...+1√n>√n Statement 2: √n(n+1)<n+1

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: $\sqrt{n (n + 1)} < n + 1$