Question

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

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

Answer 1

Answer: (D)

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

$P\left(n\right)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}$
$P\left(2\right)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2}$
Let us assume that $P\left(k\right)$
$=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}>\sqrt{k}$ is true
$\therefore P(k+1)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$ has to be true.
$L.H.S.>\sqrt{k}+\frac{1}{\sqrt{k+1}}=\frac{\sqrt{k(k+1)}+1}{\sqrt{k+1}}$
Since $\sqrt{k(k+1)}>k(\forall k\ge 0)$
$\therefore \frac{\sqrt{k(k+1)}+1}{\sqrt{k+1}}>\frac{k+1}{\sqrt{k+1}}=\sqrt{k+1}$
Let $P\left(n\right)=\sqrt{n(n+1)}<(k+1)$
Statement-1 is correct.
$P\left(2\right)=\sqrt{2\times 3}<3$
If $P\left(k\right)=\sqrt{k(k+1)}<(k+1)$ is true
Now $P(k+1)=\sqrt{(k+1)(k+2)}<k+2$ has to be true
Since $(k+1)<k+2$
$\therefore \sqrt{(k+1)(k+2)}<(k+2)$
Hence Statement-2 is not correct explanation of Statement-1.