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 $n\ge 2,$
$\sqrt{n(n+1)}<n+1.$

• 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.

Answer 1

The correct option is C

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

$P\left(n\right)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{n}}\therefore P\left(2\right)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}>\sqrt{2}$
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,
$P(k+1)=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$ has to be true.
$L.H.S.>\sqrt{k}+\frac{1}{\sqrt{k+1}}=\frac{\sqrt{k(k+1)}+1}{\sqrt{k+1}}$

Since $\sqrt{k(k+1)}>k(\forall k\le 0),$
$\frac{\sqrt{k(k+1)}+1}{\sqrt{k+1}}>\frac{k+1}{\sqrt{k+1}}=\sqrt{k+1}$
Let $P\left(n\right)=\sqrt{n(n+1)}<n+1$
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.
$P(k+1)=\sqrt{(k+1)(k+2)}<k+2$ has to be true.
Since $k+1<k+2,$
$\sqrt{(k+1)(k+2)}<k+2$
hence, statement 2 is not a correct explanation for statement 1.