Question

The sum of the series $\dfrac{12}{2!}+\dfrac{28}{3!}+\dfrac{50}{4!}+\dfrac{78}{5!}+.........$ is:
(1) $e$
(2) $4e$
(3) $3e$
(4) $5e$

Answer 1

Hint: Here in this question we have been asked to evaluate the value of the expression $\dfrac{12}{2!}+\dfrac{28}{3!}+\dfrac{50}{4!}+\dfrac{78}{5!}+.........$ . For answering this question we will find the value of the general term of the expression and then find its summation and we will use the following expansion ${{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{x}{2!}+\dfrac{x}{3!}+........$ in between the process.

Complete step-by-step solution:
Now considering from the question we have been asked to evaluate the value of the expression $\dfrac{12}{2!}+\dfrac{28}{3!}+\dfrac{50}{4!}+\dfrac{78}{5!}+.........$ .
Now we will evaluate the general term expression for the given series. In the denominator it is increasing by 1 just so we can say that the denominator will be $\left( n+1 \right)!$ .
In the numerator we have $12,28,50,78,....$ we can write this as $12,12+16,28+22,50+28,....$ this implies that we are adding the preceding term to the number 16 and adding 6 to it. Here the third term can be given as $12+16+16+6$ . From this pattern we can say that the ${{n}^{th}}$ term of this will be $12+\left( \dfrac{n-1}{2} \right)\left( 2\left( 16 \right)+\left( n-2 \right)6 \right)$ . This can be simplified as
$\begin{align}
  & \Rightarrow 12+16\left( n-1 \right)+3\left( n-1 \right)\left( n-2 \right) \\
 & \Rightarrow 12+16n-16+3{{n}^{2}}+6-9n \\
 & \Rightarrow 3{{n}^{2}}+7n+2 \\
\end{align}$
Hence we can say that the ${{n}^{th}}$ term of the given series will be $\dfrac{3{{n}^{2}}+7n+2}{\left( n+1 \right)!}$ .
Now we will simplify this as
$\begin{align}
  & \Rightarrow \dfrac{3n\left( n+1 \right)+4\left( n+1 \right)-2}{\left( n+1 \right)!} \\
 & \Rightarrow \dfrac{3}{\left( n-1 \right)!}+\dfrac{4}{n!}-\dfrac{2}{\left( n+1 \right)!} \\
\end{align}$
Now we will apply the summation for this $\Rightarrow 3\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}+4\sum\limits_{n=1}^{\infty }{\dfrac{1}{n!}}-2\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n+1 \right)!}}$ .
From the basic concepts we know the following expansion ${{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{x}{2!}+\dfrac{x}{3!}+........$ .
By substituting $x=1$ we will have $e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+........$.
 From this we can say that $\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}=e$ , $\sum\limits_{n=1}^{\infty }{\dfrac{1}{n!}}=e-1$ and $\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n+1 \right)!}}=e-2$ .
By using this identities we will have
$\begin{align}
  & \Rightarrow 3\left( e \right)+4\left( e-1 \right)-2\left( e-2 \right) \\
 & \Rightarrow 5e \\
\end{align}$
Therefore we can conclude that the sum of the given series $\dfrac{12}{2!}+\dfrac{28}{3!}+\dfrac{50}{4!}+\dfrac{78}{5!}+.........$ will be given as $5e$ . Hence we will mark the option “4” as correct.

Note: For answering questions of this type more practice is needed because the process of solving does not strike us for the first time because we can’t deduct the pattern involved in the terms. Like here generally we will not find any pattern in between the terms in the numerator but by clearing, observing and evaluating we can write the general term.