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

Hint: 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 \[{{\text{H}}_n}\].