Question

If ${{\text{H}}_n} = 1 + \dfrac{1}{2} + ............ + \dfrac{1}{n}$ , then the value of ${S_n} = 1 + \dfrac{3}{2} + \dfrac{5}{3}............ + \dfrac{{(2n - 1)}}{n}$ is
$\left( 1 \right){\text{ }}{{\text{H}}_n} + 2n$
$\left( 2 \right){\text{ n - 1 + }}{{\text{H}}_n}$
$\left( 3 \right){\text{ }}{{\text{H}}_n} - 2n$
$\left( 4 \right){\text{ }}2n - {{\text{H}}_n}$

Answer 1

Just add $1$ and $- 1$ in the numerators of the given equation of ${{\text{S}}_n}$ for the separation of terms to find the find the value of ${{\text{S}}_n}$ . Then while further solving the equation, at a particular point you will find that one of your terms is equal to the value of ${{\text{H}}_n}$ . These are the major steps for the startup of the solution. After solving the rest of the solution we are able to find the value of ${{\text{S}}_n}$.

Complete step by step answer:
The given equations are ${{\text{H}}_n} = 1 + \dfrac{1}{2} + ............ + \dfrac{1}{n}$ ------- $\left( {\text{i}} \right)$
and ${S_n} = 1 + \dfrac{3}{2} + \dfrac{5}{3}............ + \dfrac{{(2n - 1)}}{n}$ --------- $\left( {{\text{ii}}} \right)$
In the equation $\left( {{\text{ii}}} \right)$ add and subtract the number $1$ in the numerator of all terms. By doing this we get
${S_n} = \left( {1 - 1 + 1} \right) + \left( {\dfrac{{3 - 1 + 1}}{2}} \right) + \left( {\dfrac{{5 - 1 + 1}}{3}} \right) + ............. + \left( {\dfrac{{2n - 1 - 1 + 1}}{n}} \right)$
$= 1 + \left( {\dfrac{{2 + 1}}{2}} \right) + \left( {\dfrac{{4 + 1}}{3}} \right) + ............. + \left( {\dfrac{{\left( {2n - 2} \right) + 1}}{n}} \right)$
On solving further we get
$= 1 + \left( {\dfrac{2}{2} + \dfrac{1}{2}} \right) + \left( {\dfrac{4}{3} + \dfrac{1}{3}} \right) + ............. + \left( {\dfrac{{2n - 2}}{n} + \dfrac{1}{n}} \right)$
$= 1 + \dfrac{2}{2} + \dfrac{1}{2} + \dfrac{4}{3} + \dfrac{1}{3} + ............. + \dfrac{{2\left( {n - 1} \right)}}{n} + \dfrac{1}{n}$
On separating the terms with numerator $1$ from the other terms,
$=$ $\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ......... + \dfrac{1}{n}} \right)$ $+$ $\left( {\dfrac{2}{2} + \dfrac{4}{3} + ............. + \dfrac{{2\left( {n - 1} \right)}}{n}} \right)$
From the equation $\left( {\text{i}} \right)$ , we can write $\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ......... + \dfrac{1}{n}} \right)$ $= {\text{ }}{{\text{H}}_n}$
$= {{\text{H}}_n} + \left( {\dfrac{2}{2} + \dfrac{4}{3} + .............. + \dfrac{{2\left( {n - 1} \right)}}{n}} \right)$
On taking $2$ common from the term $\left( {\dfrac{2}{2} + \dfrac{4}{3} + .............. + \dfrac{{2\left( {n - 1} \right)}}{n}} \right)$ we get
$= {{\text{H}}_n} + 2\left( {\dfrac{1}{2} + \dfrac{2}{3} + .............. + \dfrac{{\left( {n - 1} \right)}}{n}} \right)$
On making the numerators of the term $\left( {\dfrac{1}{2} + \dfrac{2}{3} + .............. + \dfrac{{\left( {n - 1} \right)}}{n}} \right)$ of the form $\left( {n - 1} \right)$ we get
$= {{\text{H}}_n} + 2\left( {\dfrac{{2 - 1}}{2} + \dfrac{{3 - 1}}{3} + .............. + \dfrac{{\left( {n - 1} \right)}}{n}} \right)$
Again the separate the terms from $\left( {\dfrac{{2 - 1}}{2} + \dfrac{{3 - 1}}{3} + .............. + \dfrac{{\left( {n - 1} \right)}}{n}} \right)$ ,
$= {{\text{H}}_n} + 2\left[ {\left( {\dfrac{2}{2} + \dfrac{3}{3} + ........... + \dfrac{n}{n}} \right) + \left( { - \dfrac{1}{2} - \dfrac{1}{3} - .......... - \dfrac{1}{n}} \right)} \right]$
In the term $\left( {\dfrac{2}{2} + \dfrac{3}{3} + .......... + \dfrac{n}{n}} \right)$ the first term i.e., $1$ is missing and therefore we can write this term as $n - 1$ . Also from the term $\left( { - \dfrac{1}{2} - \dfrac{1}{3} - ............ - \dfrac{1}{n}} \right)$ , take minus sign common.
$= {{\text{H}}_n} + 2\left[ {n - 1 - \left( {\dfrac{1}{2} + \dfrac{1}{3} + ............ + \dfrac{1}{n}} \right)} \right]$
As it is given that ${{\text{H}}_n} = 1 + \dfrac{1}{2} + ............ + \dfrac{1}{n}$. From this we have $\dfrac{1}{2} + \dfrac{1}{3} + .......... + \dfrac{1}{n}$ $= {{\text{H}}_n} - 1$. Therefore the above equation becomes
$= {{\text{H}}_n} + 2\left[ {n - 1 - \left( {{{\text{H}}_n} - 1} \right)} \right]$
$= {{\text{H}}_n} + 2\left[ {n - 1 - {{\text{H}}_n} + 1} \right]$
$1$ and $- 1$ will cancel out . So,
$= {{\text{H}}_n} + 2\left[ {n - {{\text{H}}_n}} \right]$
$= {{\text{H}}_n} + 2n - 2{{\text{H}}_n}$
Further simplifying we get
$= 2n - {{\text{H}}_n}$
Therefore the correct option is $(4)$ that is $2n - {{\text{H}}_n}$.

Note:
Remember all the properties we use in the question; this will also help you in solving other questions if needed. Basically, ${{\text{S}}_n}$ represents the sum of n terms of an A.P. It is important to check while solving the problem where you should put the value of ${{\text{H}}_n}$.