Question

Solve the series $9 + \dfrac{{16}}{{2!}} + \dfrac{{27}}{{3!}} + \dfrac{{42}}{{4!}} + .....\infty $
1) $11e - 4$
2) $11e - 6$
3) $10e + 5$
4) $3e + 4$

Answer 1

Hint: First we are to deduce the given series in the form of a particular series of sum of terms. Then we are to operate the given series of terms to get the ${n^{th}}$ term of the sequence. From the ${n^{th}}$ term we can deduce the ${n^{th}}$ term of the required sequence. Then, we can obtain the sum of the ${n^{th}}$ terms of the sequence to obtain the series of sum of the terms. On calculating we can find the required answer.

Complete step-by-step solution:
The given series is, $9 + \dfrac{{16}}{{2!}} + \dfrac{{27}}{{3!}} + \dfrac{{42}}{{4!}} + .....\infty $.
So, let, $S = 9 + \dfrac{{16}}{{2!}} + \dfrac{{27}}{{3!}} + \dfrac{{42}}{{4!}} + .....\infty $.
We can write it as,
$ \Rightarrow S = \dfrac{9}{{1!}} + \dfrac{{16}}{{2!}} + \dfrac{{27}}{{3!}} + \dfrac{{42}}{{4!}} + .....\infty $.
And, let
${S_1} = 9 + 16 + 27 + 42 + .... + {t_n} - - - \left( 1 \right)$
${S_1} = \;\; 9 + 16 + 27 + 42 + .... + {t_n} - - - \left( 2 \right)$
Now, $\left( 1 \right) - \left( 2 \right)$, we get,
$0 = 9 + 7 + 11 + 15 + .... - {t_n}$
Adding ${t_n}$ on both the sides, we get,
$ \Rightarrow {t_n} = 9 + 7 + 11 + 15 + ....$
\[ \Rightarrow {t_n} = 9 + \left[ {7 + 11 + 15 + .... + {\text{upto }}\left( {n - 1} \right){\text{terms}}} \right]\]
Now, the sum of $n$ terms of an AP, is $\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where, $a = $first term of sequence, $d = $ common difference between terms and $n = $ number of terms.
Applying the formula of sum of AP, we get,
$ \Rightarrow {t_n} = 9 + \dfrac{{\left( {n - 1} \right)}}{2}\left\{ {2\left( 7 \right) + \left( {n - 1 - 1} \right)4} \right\}$
[First term $ = 7$ and Common difference $ = 4$]
$ \Rightarrow {t_n} = 9 + \dfrac{{\left( {n - 1} \right)}}{2}\left\{ {14 + \left( {n - 2} \right)4} \right\}$
Opening the brackets,
$ \Rightarrow {t_n} = 9 + \dfrac{{\left( {n - 1} \right)}}{2}\left\{ {14 + 4n - 8} \right\}$
Adding up the terms,
$ \Rightarrow {t_n} = 9 + \dfrac{{\left( {n - 1} \right)}}{2}\left\{ {6 + 4n} \right\}$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow {t_n} = 9 + \left( {n - 1} \right)\left( {2n + 3} \right)$
So, ${n^{th}}$ term of the given series $S = 9 + \dfrac{{16}}{{2!}} + \dfrac{{27}}{{3!}} + \dfrac{{42}}{{4!}} + .....\infty $ is,
${T_n} = \dfrac{{9 + \left( {n - 1} \right)\left( {2n + 3} \right)}}{{n!}}$
Opening the brackets and multiplying, we get,
$ \Rightarrow {T_n} = \dfrac{{9 + (2{n^2} + n - 3)}}{{n!}}$
$ \Rightarrow {T_n} = \dfrac{{2{n^2} + n + 6}}{{n!}}$
Diving each term in numerator by the denominator, we get,
$ \Rightarrow {T_n} = \dfrac{{2{n^2}}}{{n!}} + \dfrac{n}{{n!}} + \dfrac{6}{{n!}}$
We know, $n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right).....1 = n\left( {n - 1} \right)!$
So, on applying this, we get,
$ \Rightarrow {T_n} = \dfrac{{2{n^2}}}{{n\left( {n - 1} \right)!}} + \dfrac{n}{{n\left( {n - 1} \right)!}} + \dfrac{6}{{n!}}$
$ \Rightarrow {T_n} = \dfrac{{2n}}{{\left( {n - 1} \right)!}} + \dfrac{1}{{\left( {n - 1} \right)!}} + \dfrac{6}{{n!}}$
We can write $\left( {2n} \right)$ as $\left( {2n - 2 + 2} \right)$, applying this, we get,
$ \Rightarrow {T_n} = \dfrac{{2n - 2 + 2}}{{\left( {n - 1} \right)!}} + \dfrac{1}{{\left( {n - 1} \right)!}} + \dfrac{6}{{n!}}$
$ \Rightarrow {T_n} = \dfrac{{2n - 2}}{{\left( {n - 1} \right)!}} + \dfrac{2}{{\left( {n - 1} \right)!}} + \dfrac{1}{{\left( {n - 1} \right)!}} + \dfrac{6}{{n!}}$
Applying the property of factorials again, we get,
$ \Rightarrow {T_n} = \dfrac{{2\left( {n - 1} \right)}}{{\left( {n - 1} \right)\left( {n - 2} \right)!}} + \dfrac{2}{{\left( {n - 1} \right)!}} + \dfrac{1}{{\left( {n - 1} \right)!}} + \dfrac{6}{{n!}}$
Now, simplifying, we get,
$ \Rightarrow {T_n} = \dfrac{2}{{\left( {n - 2} \right)!}} + \dfrac{3}{{\left( {n - 1} \right)!}} + \dfrac{6}{{n!}}$
Now, to find the sum of the series,
$S = \sum {{T_n}} $
$ \Rightarrow S = \sum {\left[ {\dfrac{2}{{\left( {n - 2} \right)!}} + \dfrac{3}{{\left( {n - 1} \right)!}} + \dfrac{6}{{n!}}} \right]} $
Simplifying the expression,
$ \Rightarrow S = \sum {\dfrac{2}{{\left( {n - 2} \right)!}} + \sum {\dfrac{3}{{\left( {n - 1} \right)!}} + \sum {\dfrac{6}{{n!}}} } } $
$ \Rightarrow S = 2.\sum {\dfrac{1}{{\left( {n - 2} \right)!}} + 3.\sum {\dfrac{1}{{\left( {n - 1} \right)!}} + 6.\sum {\dfrac{1}{{n!}}} } } $
Now, simplifying the summation, we get,
$ \Rightarrow S = 2\left( {e} \right)+ 3e + 6\left( {e - 1} \right)$
$ \Rightarrow S = 11e -6 $
Therefore, the correct option of the series is $11e - 6$, correct option is 2.

Note: The sum of series has many aspects like whether the sum will converge to a finite value or not, what is the limit point of the series. Sometimes, a sum tends to appear to a number, but it will never be able to reach the number, such a series is said to be an infinite GP series, in which the GP tends to reach a particular number, but it will never reach the number.