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

The sum of the series 122!+283!+504!+785!+.........\dfrac{12}{2!}+\dfrac{28}{3!}+\dfrac{50}{4!}+\dfrac{78}{5!}+......... is:
(1) ee
(2) 4e4e
(3) 3e3e
(4) 5e5e

Explanation

Solution

Here in this question we have been asked to evaluate the value of the expression 122!+283!+504!+785!+.........\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 ex=1+x1!+x2!+x3!+........{{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 122!+283!+504!+785!+.........\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 (n+1)!\left( n+1 \right)! .
In the numerator we have 12,28,50,78,....12,28,50,78,.... we can write this as 12,12+16,28+22,50+28,....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+612+16+16+6 . From this pattern we can say that the nth{{n}^{th}} term of this will be 12+(n−12)(2(16)+(n−2)6)12+\left( \dfrac{n-1}{2} \right)\left( 2\left( 16 \right)+\left( n-2 \right)6 \right) . This can be simplified as
⇒12+16(n−1)+3(n−1)(n−2) ⇒12+16n−16+3n2+6−9n ⇒3n2+7n+2 \begin{aligned} & \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{aligned}
Hence we can say that the nth{{n}^{th}} term of the given series will be 3n2+7n+2(n+1)!\dfrac{3{{n}^{2}}+7n+2}{\left( n+1 \right)!} .
Now we will simplify this as
⇒3n(n+1)+4(n+1)−2(n+1)! ⇒3(n−1)!+4n!−2(n+1)! \begin{aligned} & \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{aligned}
Now we will apply the summation for this ⇒3∑n=1∞1(n−1)!+4∑n=1∞1n!−2∑n=1∞1(n+1)!\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 ex=1+x1!+x2!+x3!+........{{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{x}{2!}+\dfrac{x}{3!}+........ .
By substituting x=1x=1 we will have e=1+11!+12!+13!+........e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+.........
From this we can say that ∑n=1∞1(n−1)!=e\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}=e , ∑n=1∞1n!=e−1\sum\limits_{n=1}^{\infty }{\dfrac{1}{n!}}=e-1 and ∑n=1∞1(n+1)!=e−2\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n+1 \right)!}}=e-2 .
By using this identities we will have
⇒3(e)+4(e−1)−2(e−2) ⇒5e \begin{aligned} & \Rightarrow 3\left( e \right)+4\left( e-1 \right)-2\left( e-2 \right) \\\ & \Rightarrow 5e \\\ \end{aligned}
Therefore we can conclude that the sum of the given series 122!+283!+504!+785!+.........\dfrac{12}{2!}+\dfrac{28}{3!}+\dfrac{50}{4!}+\dfrac{78}{5!}+......... will be given as 5e5e . 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.