Question

The sum of the series $1+\dfrac{{{1}^{2}}+{{2}^{2}}}{2!}+\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}{3!}+\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}}{4!}+..........$ is
(a) $3e$
(b) $\dfrac{17}{6}e$
(c) $\dfrac{13}{6}e$
(d) $\dfrac{19}{6}e$

Answer 1

Hint:Observe the given series and write the general term (or nth term) of the series and apply the summation to the nth term. Use the sum of special series of adding square of natural numbers, given as
${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+..............{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
And use the expansion of exponential function i.e. ${{e}^{x}}$ , given as
${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+...........$
Put $x=1$ to the expansion to get the sum of the series given in the problem.

Complete step-by-step answer:
Here, we have to find the sum of the series
$1+\dfrac{{{1}^{2}}+{{2}^{2}}}{2!}+\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}{3!}+\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}}{4!}+..........$
Let the sum of the above series is ‘s’. So, we get
s = $1+\dfrac{{{1}^{2}}+{{2}^{2}}}{2!}+\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}{3!}+\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}}{4!}+..........\text{ }.............\left( i \right)$
We can represent the above series by writing the general term or nth term of the series and applying the summation to it. As the series ‘s’ has no end, it means we have to find the sum upto infinite.
Now, we can observe that first term has only one term in numerator $\left( \dfrac{{{1}^{2}}}{1!} \right)$ i.e ${{1}^{2}}$ and 1! in the denominator, and the second term has a sum of two terms in the numerator i.e. ${{1}^{2}}+{{2}^{2}}$ and 2! In the denominator and similarly third term has sum of three terms in the numerator i.e. ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}$ and 3! in the denominator and hence, series goes on i.e. denominator has sequence of ${{1}^{2}},{{1}^{2}}+{{2}^{2}},{{1}^{2}}+{{2}^{2}}+{{3}^{2}},..........$ and denominator has sequence of 1!, 2!, 3!, 4!.................It means the nth terms has sum of ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+........+{{n}^{2}}$ in the numerator and the term n! in the denominator. So, nth term$\left( {{T}_{n}} \right)$ can be given as
${{T}_{n}}=\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+.........{{n}^{2}}}{n!}.............(ii)$
So, ${{S}_{n}}$(sum upto n terms) can be given as
${{S}_{n}}=\sum\limits_{n=1}^{n}{\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.........{{n}^{2}}}{n!}..........(iii)}$
Now, we can get S, by applying $n\to \infty$ to the series written in equation (iii), as the series ‘S’ has infinite terms. So, we can write series S from the equation (iii) as
${{S}_{n}}=\sum\limits_{n=1}^{\infty }{\dfrac{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.........{{n}^{2}}}{n!}..........(iv)}$
Now, we know the series in the numerator of equation (iv) is specials series of sum of square of natural numbers, which is given as
${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+...........{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
So equation (iv) can be written as
$\begin{aligned} & S=\sum\limits_{n=1}^{\infty }{\dfrac{\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}{n!}} \\\ & S=\sum\limits_{n=1}^{\infty }{\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6\left( n! \right)}...........\left( v \right)} \\\ \end{aligned}$
We know n! can also be written as
$n!=n\left( n-1 \right)!............\left( vi \right)$
So, we get value of s using equation (v) and (vi) as
$\begin{aligned} & S=\sum\limits_{n=1}^{\infty }{\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6n\left( n-1 \right)!}} \\\ & \Rightarrow S=\sum\limits_{n=1}^{\infty }{\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{6\left( n-1 \right)!}} \\\ \end{aligned}$
We can write $\left( n+1 \right)$ as $\left( \left( n-1 \right)+2 \right)$ as well. So, we get
$\begin{aligned} & S=\sum\limits_{n=1}^{\infty }{\dfrac{\left( \left( n-1 \right)+2 \right)\left( 2n+1 \right)}{6\left( n-1 \right)!}} \\\ & \Rightarrow S=\sum\limits_{n=1}^{\infty }{\dfrac{\left( n-1 \right)\left( 2n+1 \right)+2\left( 2n+1 \right)}{6\left( n-1 \right)!}} \\\ \end{aligned}$
Now, we can divide both terms in the numerator individually. So, we get
$S=\sum\limits_{n=1}^{\infty }{\dfrac{\left( 2n+1 \right)\left( n-1 \right)}{6\left( n-1 \right)!}+\dfrac{2\left( 2n+1 \right)}{6\left( n-1 \right)!}}$
We can write $\left( n-1 \right)!$ in the first fraction as
$\left( n-1 \right)!=\left( n-1 \right)\left( n-2 \right)!$
So, we get
$\begin{aligned} & S=\sum\limits_{n=1}^{\infty }{\dfrac{\left( 2n+1 \right)\left( n-1 \right)}{6\left( n-1 \right)\left( n-2 \right)!}+\dfrac{2\left( 2n+1 \right)}{6\left( n-1 \right)!}} \\\ & \Rightarrow S=\sum\limits_{n=1}^{\infty }{\left[ \dfrac{\left( 2n+1 \right)}{6\left( n-2 \right)!}+\dfrac{\left( 2n+1 \right)}{3\left( n-1 \right)!} \right]} \\\ \end{aligned}$
Now, we can write (2n + 1) of the first fraction as
2n + 1 = 2n – 4 + 5 = 2(n – 2) + 5
And similarly, we can write (2n + 1) of the second fraction as
2n + 1 = 2n – 2 + 3 = 2(n – 1) + 3
So, we can write ‘S’ as

& S=\sum\limits_{n=1}^{\infty }{\left[ \dfrac{2\left( n-2 \right)+5}{6\left( n-2 \right)!}+\dfrac{2\left( n-1 \right)+3}{3\left( n-1 \right)!} \right]} \\\ & S=\sum\limits_{n=1}^{\infty }{\left[ \dfrac{2\left( n-2 \right)}{6\left( n-2 \right)!}+\dfrac{5}{6\left( n-2 \right)!}+\dfrac{2\left( n-1 \right)}{3\left( n-1 \right)!}+\dfrac{3}{3\left( n-1 \right)!} \right]} \\\ & S=\sum\limits_{n=1}^{\infty }{\left[ \dfrac{\left( n-2 \right)}{3\left( n-2 \right)!}+\dfrac{5}{6\left( n-2 \right)!}+\dfrac{2\left( n-1 \right)}{3\left( n-1 \right)!}+\dfrac{1}{\left( n-1 \right)!} \right]} \\\ \end{aligned}$$ Now, we can write (n - 2)! In the first fraction as $\left( n-2 \right)!=\left( n-2 \right)\left( n-3 \right)!$ and similarly, (n – 1)! in the third fraction as $\left( n-1 \right)!=\left( n-1 \right)\left( n-2 \right)!$ $$\begin{aligned} & S=\sum\limits_{n=1}^{\infty }{\left[ \dfrac{\left( n-2 \right)}{3\left( n-2 \right)\left( n-3 \right)!}+\dfrac{5}{6\left( n-2 \right)!}+\dfrac{2\left( n-1 \right)}{3\left( n-1 \right)\left( n-2 \right)!}+\dfrac{1}{\left( n-1 \right)!} \right]} \\\ & S=\sum\limits_{n=1}^{\infty }{\left[ \dfrac{1}{3\left( n-3 \right)!}+\dfrac{5}{6\left( n-2 \right)!}+\dfrac{2}{3\left( n-2 \right)!}+\dfrac{1}{\left( n-1 \right)!} \right]} \\\ \end{aligned}$$ Now, we can apply summation to all the terms written inside the bracket $$S=\dfrac{1}{3}\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-3 \right)!}}+\dfrac{5}{6}\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-2 \right)!}}+\dfrac{2}{3}\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-2 \right)!}}+\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}$$ Now, we can observe the four series as written above. We can get the first series should start from n = 2 and third from n = 2 as well and the last one should start from n = 1 because r! cannot take negative value of r. So, we get $$S=\dfrac{1}{3}\sum\limits_{n=3}^{\infty }{\dfrac{1}{\left( n-3 \right)!}}+\dfrac{5}{6}\sum\limits_{n=2}^{\infty }{\dfrac{1}{\left( n-2 \right)!}}+\dfrac{2}{3}\sum\limits_{n=2}^{\infty }{\dfrac{1}{\left( n-2 \right)!}}+\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}.......(vii)$$ As we know expansion ${{e}^{x}}$ can be given as ${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+.........$ Put $x=1,$ to the above equation, we get $\begin{aligned} & e=1+1+\dfrac{{{1}^{2}}}{2!}+\dfrac{{{1}^{3}}}{3!}+\dfrac{{{1}^{4}}}{4!}+......... \\\ & e=1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+.........\text{ }........\left( viii \right) \\\ \end{aligned}$ Now let us expand all the series involved in equation (vii) as $s=\dfrac{1}{3}\left[ \dfrac{1}{0!}+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+... \right]+\dfrac{5}{6}\left[ \dfrac{1}{0!}+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+... \right]+\dfrac{2}{3}\left[ \dfrac{1}{0!}+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+... \right]+\left[ \dfrac{1}{0!}+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+... \right]$ We know 0! = 1, So we get, $s=\dfrac{1}{3}\left[ 1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+... \right]+\dfrac{5}{6}\left[ 1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+... \right]+\dfrac{2}{3}\left[ 1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+... \right]+\left[ 1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+... \right]$ Now, we can use equation (viii) and replace the term by e. So, we get $\begin{aligned} & s=\dfrac{1}{3}e+\dfrac{5}{6}e+\dfrac{2}{3}e+1e \\\ & s=\dfrac{2e+5e+4e+6e}{6} \\\ & s=\dfrac{17e}{6} \\\ \end{aligned}$ So, option (B) is correct answer Note: One may go wrong if he/she solves all the terms given in the series and then try to observe some series. He /she will not be able to notice any particular series. So, the given series is already following a particular order, observe it and hence solve it. One may try to think of any special series related to A.P, G.P or H.P, but the given series is an expansion of ${{e}^{x}}$ so, be clear with the expansions of the functions as well. So, remember those things as well. Don’t confuse with the terms $\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-3 \right)!}\text{ (or) }\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-2 \right)!}}}$ .Here , we cannot put those values of ‘n’ which will give negative value of $\dfrac{1}{n-2}$ or $\dfrac{1}{n-3}$ . So, be careful and don’t confuse yourself. Always remember that $r!$ cannot take negative values