Question

Statement – I : For every natural number ‘n’, at $n{\text{ }} \geqslant {\text{ }}2$, given that : $\dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}$ where, $\dfrac{1}{{\sqrt n }} > \sqrt n$.
Statement – II : For every natural number ‘n’, at $n{\text{ }} \geqslant {\text{ }}2$, given that : $\sqrt {n\left( {n + 1} \right)}$ where, $n < n + 1$.
State whether the statements are right or wrong.
(a) Only Statement – I is right.
(b) Neither statement – I is right nor statement – II.
(c) Only Statement – II is right.
(d) Both statements are right but statement – II is not the right explanation of statement – I.

Answer 1

The given problem revolves around the concepts algebraic terms considering the given conditions correct and substituting these respective conditions or parameters, etc. the desired conclusion is obtained. Just put the $n{\text{ }} \geqslant {\text{ }}2$ parameter in both the statements that is $\dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}$ and $\sqrt {n\left( {n + 1} \right)}$ for $\dfrac{1}{{\sqrt n }} > \sqrt n$ and $n < n + 1$ respectively, then draw the required conclusion.

Complete answer:
(For Statement – I),
Since, considering the statement – I, that is
$\dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}$
As a result, from the given condition at $n{\text{ }} \geqslant {\text{ }}2$, we get
$\dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} > \sqrt 2$
Solving the equation mathematically, we get
$\dfrac 1 + 0.7071 > 1.4142 \\\ 1.7071 > 1.4142 \\\$
Hence, it seems that the given condition at $n{\text{ }} \geqslant {\text{ }}2$ is correct..!
As a result, let us assume the condition $\dfrac{1}{{\sqrt n }} > \sqrt n$ at $n{\text{ }} \geqslant {\text{ }}2$, we get
(To solve the equation assume $n = k$)
$\dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt k }} > \sqrt k$ … (i)
Similarly,
Considering the condition included in statement – II, that is $n < n + 1$, we get
(To solve the equation assume $n = k + 1$)
Hence, equation (i) becomes
$\dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt k }} + \dfrac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1}$ … (ii)
(For Statement – II),
Since, considering the statement – I, that is
$\sqrt {n\left( {n + 1} \right)}$
As a result, let us assume the condition $\sqrt {n\left( {n + 1} \right)}$ at $n{\text{ }} \geqslant {\text{ }}2$, we get
(To solve the equation assume $n = k$)
$\sqrt {k\left( {k + 1} \right)} < k + 1$
Solving the terms mathematically, we get
$\sqrt k \sqrt {k + 1} < \sqrt {k + 1} \sqrt {k + 1}$
$\sqrt k < \sqrt {k + 1}$
But, we have given
For $k \geqslant 2$,
$\sqrt {k + 1} > \sqrt k$
Hence, the equation becomes
$\dfrac{{\sqrt k }}{{\sqrt {k + 1} }} < 1$
Multiplying by $\sqrt k$, we get
$\dfrac{k}{{\sqrt {k + 1} }} < \sqrt k$
Mathematically equation can be written as,
$\dfrac{{\left( {k + 1} \right) - 1}}{{\sqrt {k + 1} }} < \sqrt k$
Separating the numerator and denominator, we get
$\sqrt {k + 1} - \dfrac{1}{{\sqrt {k + 1} }} < \sqrt k$
$\sqrt {k + 1} < \sqrt k + \dfrac{1}{{\sqrt {k + 1} }}$ … (iii)
Hence, from (i), (ii) and (iii),
It seems that,
$\dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt k }} + \dfrac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1}$
Both the statements are true for the given parameter $n{\text{ }} \geqslant {\text{ }}2$ but statement – II is not the right explanation for statement – I respectively.
Therefore, option (D) is the correct answer.

Note:
One must be able to know the basic fundamentals of algebraic terms such as solving equations by substituting the given parameters, conditions, etc. Also, need to analyze the conditions at two different situations given in the problem, so as to be sure of our final answer.