Question

Find the sum of n terms of the series
$\dfrac{2}{{1.2}} + \dfrac{5}{{2.3}}.2 + \dfrac{{10}}{{3.4}}{2^2} + \dfrac{{17}}{{4.5}}{.2^3} + ......$

Answer 1

In this particular type of question use the concept that first find out the ${n^{th}}$ term of the given series by observing the trend it follows then separate its ${n^{th}}$ terms in the terms of denominator and then write its, ${T_1},{T_2},{T_3}..............{T_{n - 1}},{T_n}$ terms and add them so that it cancel out most of the terms so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given series is
$\dfrac{2}{{1.2}} + \dfrac{5}{{2.3}}.2 + \dfrac{{10}}{{3.4}}{2^2} + \dfrac{{17}}{{4.5}}{.2^3} + ......$
In the above series first find out the ${n^{th}}$ term of the above series.
So the numerator of the above series is
2, 5, 10, 17,......
So as we see carefully, it will follow the trend of $\left( {{n^2} + 1} \right)$ , where n = 1, 2, 3, ...
Now the numerator is multiplied by $1,2,{2^2},{2^3},.....$
So this will follow the G.P with common ratio 2.
So the ${n^{th}}$ term of this series is ${r^{n - 1}}$
So the ${n^{th}}$ term of $1,2,{2^2},{2^3},.....$ is ${2^{n - 1}}$ , where n = 1, 2, 3, ...
So over all numerator of the given series follow the trend of $\left( {{n^2} + 1} \right){2^{n - 1}}$
Now the denominator of the given series is (1.2, 2.3, 3.4, 4.5........)
So as we carefully, it will follow the trend of $n\left( {n + 1} \right)$ , where n = 1, 2, 3, ...
So the ${n^{th}}$ term of the given series is
$\Rightarrow {T_n} = \dfrac{{\left( {{n^2} + 1} \right){2^{n - 1}}}}{{n\left( {n + 1} \right)}}$
Now add and subtract by 1 in numerator of $\dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}}$ term we have,
$\Rightarrow \dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}} = \dfrac{{\left( {{n^2} - 1 + 1 + 1} \right)}}{{n\left( {n + 1} \right)}}$
$\Rightarrow \dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}} = \dfrac{{\left( {{n^2} - 1} \right)}}{{n\left( {n + 1} \right)}} + \dfrac{{\left( 2 \right)}}{{n\left( {n + 1} \right)}}$
$\Rightarrow \dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}} = \dfrac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{n\left( {n + 1} \right)}} + \dfrac{{\left( 2 \right)}}{{n\left( {n + 1} \right)}}$
$\Rightarrow \dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}} = \dfrac{{\left( {n - 1} \right)}}{n} + \dfrac{{\left( 2 \right)}}{{n\left( {n + 1} \right)}}$
$\Rightarrow \dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}} = 1 - \dfrac{1}{n} + \dfrac{{\left( 2 \right)}}{{n\left( {n + 1} \right)}}$ , $\left[ {\because \dfrac{{\left( 2 \right)}}{{n\left( {n + 1} \right)}} = \dfrac{2}{n} - \dfrac{2}{{n + 1}}} \right]$
$\Rightarrow \dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}} = 1 - \dfrac{1}{n} + \dfrac{2}{n} - \dfrac{2}{{n + 1}}$
$\Rightarrow \dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}} = 1 - \dfrac{2}{{n + 1}} + \dfrac{1}{n}$
So this is also written as,
$\Rightarrow \dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}} = \left( {2 - \dfrac{2}{{n + 1}}} \right) - \left( {1 - \dfrac{1}{n}} \right)$
$\Rightarrow \dfrac{{\left( {{n^2} + 1} \right)}}{{n\left( {n + 1} \right)}} = \left( {\dfrac{{2n}}{{n + 1}}} \right) - \left( {\dfrac{{n - 1}}{n}} \right)$
So ${T_n}$ become
$\Rightarrow {T_n} = \dfrac{{\left( {{n^2} + 1} \right){2^{n - 1}}}}{{n\left( {n + 1} \right)}} = \left[ {\left( {\dfrac{{2n}}{{n + 1}}} \right) - \left( {\dfrac{{n - 1}}{n}} \right)} \right]{2^{n - 1}}$
$\Rightarrow {T_n} = \dfrac{{\left( {{n^2} + 1} \right){2^{n - 1}}}}{{n\left( {n + 1} \right)}} = \left( {\dfrac{{{2^n}.n}}{{n + 1}}} \right) - \left( {\dfrac{{{2^{n - 1}}\left( {n - 1} \right)}}{n}} \right)$
Now write ${T_1},{T_2},{T_3}..............{T_{n - 1}},{T_n}$ we have,
${T_1} = \dfrac{{2.1}}{2} - 0$
${T_2} = \left( {\dfrac{{{2^2}.2}}{3}} \right) - \left( {\dfrac{{2.1}}{2}} \right)$
${T_3} = \left( {\dfrac{{{2^3}.3}}{4}} \right) - \left( {\dfrac{{{2^2}.2}}{3}} \right)$
${T_3} = \left( {\dfrac{{{2^4}.4}}{5}} \right) - \left( {\dfrac{{{2^3}.3}}{4}} \right)$
.
.
.
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${T_{n - 1}} = \left( {\dfrac{{{2^{n - 1}}.\left( {n - 1} \right)}}{n}} \right) - \left( {\dfrac{{{2^{n - 2}}\left( {n - 2} \right)}}{{n - 1}}} \right)$
${T_n} = \left( {\dfrac{{{2^n}.n}}{{n + 1}}} \right) - \left( {\dfrac{{{2^{n - 1}}\left( {n - 1} \right)}}{n}} \right)$
Now add these all terms we have,
$\Rightarrow {T_1} + {T_2} + {T_3} + ... + {T_{n - 1}} + {T_n} = {S_n} = \dfrac{{{2^n}.n}}{{n + 1}} - 0$
Remaining all the terms are cancel out so we have,
$\Rightarrow {S_n} = \dfrac{{{2^n}.n}}{{n + 1}}$
So this is the required sum.

Note: Whenever we face such types of questions the key concept we have to remember is that first convert the given series into simplified form by converting its ${n^{th}}$ term into the difference of two term such that when we add all the terms by substituting n =1 ,2, 3, 4....... (n – 1), the most of the terms are canceled out and the remaining terms are the sum of the required series.