Question

Statement-$1$ : For every natural number $n\ge 2,\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>\sqrt{n}$.

Statement-$2$ : For every natural number $n\ge 2,\sqrt{n\left(n+1\right)}<n+1$.

• A. Statement-$1$ is true, Statement-$2$ is false
• B. Statement-$1$ is false, Statement-$2$ is true
• C. Statement-$1$ is true, Statement-$2$ is true; Statement-$2$ is a correct explanation for Statement-$1$
• D. Statement-$1$ is true, Statement-$2$ is true; Statement-$2$ is not a correct explanation for Statement-$1$

Answer 1

The correct option is D

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

Explanation for the correct option:

Solving by induction:

Case I:

Let $P\left(n\right):\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}>\sqrt{n}$

This statement is true for $n=2$. That is $P\left(2\right)$ is true.

Now assume that the statement is true for $n=k$. That is $P\left(k\right)$ is true.

$$\begin{gathered} \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}>\sqrt{k} \\ \Rightarrow \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k}+\frac{1}{\sqrt{k+1}} \\ \Rightarrow \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\frac{\sqrt{k\left(k+1\right)}+1}{\sqrt{k+1}} \end{gathered}$$

Since $\sqrt{k\left(k+1\right)}>k$ for all $k\ge 0$.

$$\begin{gathered} \therefore \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\frac{k+1}{\sqrt{k+1}} \\ \Rightarrow \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1} \end{gathered}$$

Therefore, $P(k+1)$ is true.

Then by mathematical induction, we can say that $P\left(n\right)$ is true for all $n\ge 2$.

So, Statement-$1$ is true.

Case II:

Let $P\left(n\right):\sqrt{n\left(n+1\right)}<n+1$

This statement is true for $n=2$. That is $P\left(2\right)$ is true.

Now assume that the statement is true for $n=k$. That is $P\left(k\right)$ is true.

$\therefore \sqrt{k(k+1)}<k+1$

Now let $n=k+1$

$$\begin{gathered} \Rightarrow LHS=\sqrt{\left(k+1\right)\left(k+2\right)} \\ <\sqrt{\left(k+2\right)(k+2)} \\ =k+2 \end{gathered}$$

Hence Statement-$2$ is true. but is not the correct explanation for the statement-$1$.

Therefore, the correct answer is option (D).