First slide

Introduction to P.M.I

Question

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

Moderate

A

Statement 1 is false, Statement 2 is true
B

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

Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1
D

Statement 1 is true, Statement 2 is false

Solution

$\begin{array}{l} P (n) = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} \\ P (2) = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} > \sqrt{2} \end{array}$
Let us assume that $P (k) = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{k}} > \sqrt{k}$ is true.
$∴ P (k + 1) = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k + 1}} > \sqrt{k + 1}$ has to be true .
$L.H.S. > \sqrt{k} + \frac{1}{\sqrt{k + 1}} = \frac{\sqrt{k (k + 1)} + 1}{\sqrt{k + 1}}$ since $\sqrt{k (k + 1)} > k (\forall k \geq 0)$
$∴ \frac{\sqrt{k (k + 1)} + 1}{\sqrt{k + 1}} > \frac{k + 1}{\sqrt{k + 1}} = \sqrt{k + 1}$
Let $P (n) = \sqrt{n (n + 1)} < n + 1$
Statement 1 is correct.
$P (2) = \sqrt{2 \times 3} < 3$
If $P (k) = \sqrt{k (k + 1)} < (k + 1)$ id true
Now $P (k + 1) = \sqrt{(k + 1) (k + 2)} < k + 2$ has to be true.
since $(k + 1) < k + 2$
$∴ \sqrt{(k + 1) (k + 2)} < (k + 2)$
Hence, Statement 2 is not a correct explanation of Statement 1.

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