Question

The sum of $\dfrac{{{1^2} \cdot 2}}{{1!}} + \dfrac{{{2^2} \cdot 3}}{{2!}} + \dfrac{{{3^2} \cdot 4}}{{3!}} + .....\infty $ is
A. $5e$
B. $3e$
C. $7e$
D. $2e$

Answer 1

Hint: The terms given in the series follows a pattern. Find the pattern and use it to find the $n$-th term of the given series. Then convert the fraction into a sum of partial fractions. Make a suitable arrangement for the given terms using the $n$-th term.

Formula Used:
Expansion of $e$ is $e = 1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....$
The factorial of $n$ is $n! = n\left( {n - 1} \right)!$

Complete step by step solution:
The given series is $\dfrac{{{1^2} \cdot 2}}{{1!}} + \dfrac{{{2^2} \cdot 3}}{{2!}} + \dfrac{{{3^2} \cdot 4}}{{3!}} + .....\infty $
Let us find the $n$-th term of the series.
The numerators of each term are products of two numbers and denominators are factorials.
The sequence of the first numbers of the numerators is ${1^2},{2^2},{3^2},.....$
So, the first number of the numerator of the $n$-th term is ${n^2}$.
The sequence of the second numbers of the numerators is $2,3,4,.....$
So, the second number of the numerator of the $n$-th term is $n + 1$.
The sequence of the denominators is $1!,2!,3!,.....$
So, the denominator of the $n$-th term is $n!$
Thus, the $n$-th term of the series is $\dfrac{{{n^2}\left( {n + 1} \right)}}{{n!}}$
Let ${T_n} = \dfrac{{{n^2}\left( {n + 1} \right)}}{{n!}}$
The factorial of $n$ is $n! = n\left( {n - 1} \right)!$
So, ${T_n} = \dfrac{{{n^2}\left( {n + 1} \right)}}{{n\left( {n - 1} \right)!}} = \dfrac{{n\left( {n + 1} \right)}}{{\left( {n - 1} \right)!}}$
Let us try to rewrite this fraction as a partial sum of fractions.
We can write $n\left( {n + 1} \right) = {n^2} + n = {n^2} + 2n - n - 2 + 2 = n\left( {n + 2} \right) - \left( {n + 2} \right) + 2 = \left( {n + 2} \right)\left( {n - 1} \right) + 2$
So, ${T_n} = \dfrac{{\left( {n + 2} \right)\left( {n - 1} \right) + 2}}{{\left( {n - 1} \right)\left( {n - 2} \right)!}}$
Separate the terms in the numerator.
${T_n} = \dfrac{{\left( {n + 2} \right)\left( {n - 1} \right) + 2}}{{\left( {n - 1} \right)\left( {n - 2} \right)!}} = \dfrac{{\left( {n + 2} \right)\left( {n - 1} \right)}}{{\left( {n - 1} \right)\left( {n - 2} \right)!}} + \dfrac{2}{{\left( {n - 1} \right)\left( {n - 2} \right)!}} = \dfrac{{n + 2}}{{\left( {n - 2} \right)!}} + \dfrac{2}{{\left( {n - 1} \right)!}} - - - - - \left( i \right)$
Again, we can write $n + 2 = n - 2 + 4$
So, ${T_n} = \dfrac{{n - 2 + 4}}{{\left( {n - 2} \right)!}} + \dfrac{2}{{\left( {n - 1} \right)!}} = \dfrac{{n - 2}}{{\left( {n - 2} \right)\left( {n - 3} \right)!}} + \dfrac{4}{{\left( {n - 2} \right)!}} + \dfrac{2}{{\left( {n - 1} \right)!}} = \dfrac{1}{{\left( {n - 3} \right)!}} + \dfrac{4}{{\left( {n - 2} \right)!}} + \dfrac{2}{{\left( {n - 1} \right)!}} - - - - - \left( {ii} \right)$
The first term is ${T_1} = \dfrac{{{1^2} \cdot 2}}{{1!}} = \dfrac{2}{1} = 2$
Putting $n = 2$ in the expression $\left( i \right)$, we get the second term given by ${T_2} = \dfrac{4}{{0!}} + \dfrac{2}{{1!}} = 4 + \dfrac{2}{{1!}}$
Putting $n = 3$ in the expression $\left( {ii} \right)$, we get the third term given by ${T_3} = \dfrac{1}{{0!}} + \dfrac{4}{{1!}} + \dfrac{2}{{2!}} = 1 + \dfrac{4}{{1!}} + \dfrac{2}{{2!}}$
Putting $n = 4$ in the expression $\left( {ii} \right)$, we get the fourth term given by ${T_4} = \dfrac{1}{{1!}} + \dfrac{4}{{2!}} + \dfrac{2}{{3!}}$
Putting $n = 5$ in the expression $\left( {ii} \right)$, we get the fifth term given by ${T_5} = \dfrac{1}{{2!}} + \dfrac{4}{{3!}} + \dfrac{2}{{4!}}$
and so on.
Adding all the terms, we get
$ \Rightarrow \left( 2 \right) + \left( {4 + \dfrac{2}{{1!}}} \right) + \left( {1 + \dfrac{4}{{1!}} + \dfrac{2}{{2!}}} \right) + \left( {\dfrac{1}{{1!}} + \dfrac{4}{{2!}} + \dfrac{2}{{3!}}} \right) + \left( {\dfrac{1}{{2!}} + \dfrac{4}{{3!}} + \dfrac{2}{{4!}}} \right) + .......$
Arrange the terms
$ \Rightarrow \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right) + 2\left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right) + 4\left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right)$
The series $\left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right)$ is the expansion of $e$.
So, the required sum is $ e + 2e + 4e = 7e$

Option ‘C’ is correct

Note: Always remember to convert an expression as a sum of partial fractions. At then end you will find a special expansion $\left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right)$, which is the expansion of $e$. Substitute $e = 1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....$ and obtain the required sum.