Question

The ${n^{(th)}}$ term of the series $\dfrac{2}{{1!}}{\text{ }} + {\text{ }}\dfrac{7}{{2!}}{\text{ }} + {\text{ }}\dfrac{{15}}{{3!}}{\text{ }} + {\text{ }}\dfrac{{26}}{{4!}}{\text{ }} + {\text{ }}...$is
$\left( 1 \right){\text{ }}\dfrac{{n\left( {3n - 1} \right)}}{{2\left( {n!} \right)}}$
$\left( 2 \right){\text{ }}\dfrac{{n\left( {3n + 1} \right)}}{{2\left( {n!} \right)}}$
$\left( 3 \right){\text{ }}\dfrac{{n\left( {3n} \right)}}{{2\left( {n!} \right)}}$
$\left( 4 \right){\text{ }}none{\text{ }}of{\text{ }}these$

Answer 1

Hint : We have to find the ${n^{(th)}}$ term of the given series . We solve this question using the concept of sum to n terms of a special series . We would simplify the given equation to form an AP and on solving and equating the equations we can find the value of ${n^{(th)}}$ term , using first term $\left( a \right)$ and common difference $\left( d \right)$ . We will put the value of a and d in the formula of the sum of n terms to find the ${n^{(th)}}$ term of the series.

Complete step-by-step answer :
For the terms of a given series to be in A.P. the common difference between the terms of the series should be the same for all the two consecutive terms of the series . The difference of the second term to the first term of the given series should be the same as that of the difference of the third term to the second term of the given series .
Given series : $\dfrac{2}{{1!}} + {\text{ }}\dfrac{7}{{2!}}{\text{ }} + {\text{ }}\dfrac{{15}}{{3!}}{\text{ }} + {\text{ }}\dfrac{{26}}{{4!}}{\text{ }} + {\text{ }} \ldots$
Here the denominator terms are $1!,{\text{ }}2!,{\text{ }}3!, \ldots$
So ,
The denominator of ${n^{(th)}}$ term = $n!$
Now ,
Consider the terms in the numerator $2,{\text{ }}7,{\text{ }}15,{\text{ }}26 \ldots ,{a_n}$
Here, ${n^{(th)}}$ term of the numerator
${a_n}{\text{ }} = {\text{ }}2{\text{ }} + {\text{ }}\left( {5{\text{ }} + {\text{ }}8{\text{ }} + {\text{ }}11 + \ldots n - 1} \right)$
[ Taking upto $\left( {n - 1} \right){\text{ }}terms$ as we have taken first term out of the bracket ]
Now, applying the formula of sum of n terms of A.P.
${s_n}{\text{ }} = \left[ {\dfrac{n}{2}} \right] \times [2a + \left( {n - 1} \right) \times d]$———(1)
since ,
$5,8,11 \ldots \left( {n - 1} \right)$ is an AP
$a{\text{ }} = {\text{ }}5$ , $d{\text{ }} = {\text{ }}3$ and $n{\text{ }} = {\text{ }}n - 1$
Substituting the values as per question in (1)
${a_n} = {\text{ }}2{\text{ }} + {\text{ }}\left[ {\dfrac{{\left( {n - 1} \right)}}{2}} \right]{\text{ }} \times {\text{ }}\left[ {10{\text{ }} + \left( {n - 2} \right)3} \right]$
${a_n}{\text{ }} = {\text{ }}2{\text{ }} + \left[ {\dfrac{{\left( {n - 1} \right)}}{2}} \right]{\text{ }} \times {\text{ }}\left[ {10 + 3n - 6} \right]$
$= \dfrac{{\left[ {4{\text{ }} + {\text{ }}\left( {n - 1} \right)\left( {3n + 4} \right)} \right]}}{2}$
$= \dfrac{{\left[ {4{\text{ }} + {\text{ }}3{n^2}{\text{ }}-{\text{ }}3n{\text{ }} + {\text{ }}4n{\text{ }}-{\text{ }}4} \right]}}{2}$
$= {\text{ }}\dfrac{{\left( {3{n^2} + n} \right)}}{2}$
$= {\text{ }}\dfrac{{n\left( {3n + 1} \right)}}{2}$
${a_n}$is the numerator of the ${n^{(th)}}$ term of the series
For the ${n^{(th)}}$ term of the series
${n^{(th)}}term$ $= \dfrac{{numerator\; of\; {n^{(th)}}term}}{{denominator \;of \;the \;{n^{(th)}}term}}$
So,
${n^{(th)}}$ term of series $= {\text{ }}\dfrac{{n\left( {3n + 1} \right)}}{{2\left( {n!} \right)}}$
Thus , the correct option is $\left( 2 \right)$
So, the correct answer is “Option 2”.

Note: We used the concept of ${n^{(th)}}$ term of AP , the formula of sum of $n$ terms of an AP, the concept of special series .
The sum of first $n$terms of some special series :
$1{\text{ }} + {\text{ }}2{\text{ }} + {\text{ }}3{\text{ }} + {\text{ }}4{\text{ }} + {\text{ }} \ldots \ldots \ldots \ldots \ldots .{\text{ }} + {\text{ }}n{\text{ }} = {\text{ }}\dfrac{{n\left( {n + 1} \right)}}{2}$
${1^2} + {2^2} + {3^3} + ...... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}$
${1^3} + {2^3} + {3^3} + ....... + {n^3} = \dfrac{{{{[n(n + 1)]}^2}}}{4}$