Question

Find \[\sum\limits_{n = 1}^\infty {\dfrac{{\left( {2{n^2} + n + 1} \right)}}{{n!}}} \] is equal to
(1) \[2e - 1\]
(2) \[2e + 1\]
(3) \[6e - 1\]
(4) \[6e + 1\]

Answer 1

Hint: To solve this question, first we will divide each term of the numerator by denominator separately and apply summation on them. After that using the formula of factorial i.e., \[n! = n\left( {n - 1} \right)!\] we will simplify the terms. After simplification we will expand the terms, then we will see that there is an expansion formed of exponential function i.e., \[{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....\] then with the help of this expansion we will simplify the terms and get the desired result.

Complete step-by-step answer:
The given expansion is, \[\sum\limits_{n = 1}^\infty {\dfrac{{\left( {2{n^2} + n + 1} \right)}}{{n!}}} \]
First of all, divide each term of the numerator by denominator separately and apply summation on them
Therefore, we get
\[\sum\limits_{n = 1}^\infty {\dfrac{{2{n^2}}}{{n!}} + } {\text{ }}\sum\limits_{n = 1}^\infty {\dfrac{n}{{n!}} + } {\text{ }}\sum\limits_{n = 1}^\infty {\dfrac{1}{{n!}}} \]
Now, using the formula \[n! = n\left( {n - 1} \right)!\] we can write above expression as,
\[\sum\limits_{n = 1}^\infty {\dfrac{{2{n^2}}}{{n\left( {n - 1} \right)!}} + } {\text{ }}\sum\limits_{n = 1}^\infty {\dfrac{n}{{n\left( {n - 1} \right)!}} + } {\text{ }}\sum\limits_{n = 1}^\infty {\dfrac{1}{{n!}}} \]
On cancelling \[n\] from the first two terms from the numerator and denominator, we get
\[\sum\limits_{n = 1}^\infty {\dfrac{{2n}}{{\left( {n - 1} \right)!}} + } {\text{ }}\sum\limits_{n = 1}^\infty {\dfrac{1}{{\left( {n - 1} \right)!}} + } {\text{ }}\sum\limits_{n = 1}^\infty {\dfrac{1}{{n!}}} \]
Now on applying the summation, we can write,
\[\left( {\dfrac{2}{{0!}} + \dfrac{{2 \cdot 2}}{{1!}} + \dfrac{{2 \cdot 3}}{{2!}} + .....} \right) + \left( {\dfrac{1}{{0!}} + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right) + \left( {\dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + .....} \right)\]
Taking common \[2\] from the first bracket, we get
\[2\left( {\dfrac{1}{{0!}} + \dfrac{2}{{1!}} + \dfrac{3}{{2!}} + .....} \right) + \left( {\dfrac{1}{{0!}} + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right) + \left( {\dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + .....} \right)\]
We know that \[0! = 1\]
Therefore, the above expression becomes,
\[2\left( {1 + \dfrac{2}{{1!}} + \dfrac{3}{{2!}} + .....} \right) + \left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right) + \left( {\dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + .....} \right)\]
As we know that \[{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ....\]
\[ \Rightarrow {e^1} = 1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....\]
So, the above expression can be written as,
\[2\left( {1 + \dfrac{2}{{1!}} + \dfrac{3}{{2!}} + .....} \right) + \left( e \right) + \left( {e - 1} \right) - - - \left( 1 \right)\]
Now, we will first solve the first bracket
Basically, we will try to make this in the form of exponential expansion,
So, \[2\left( {1 + \dfrac{2}{{1!}} + \dfrac{3}{{2!}} + .....} \right)\] can be written as,
\[2\left( {1 + \dfrac{{1 + 1}}{{1!}} + \dfrac{{1 + 2}}{{2!}} + \dfrac{{1 + 3}}{{3!}} + .....} \right)\]
which can also be written as,
\[2\left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right) + 2\left( {\dfrac{1}{{1!}} + \dfrac{2}{{2!}} + \dfrac{3}{{2!}} + .....} \right)\]
Using \[{e^1} = 1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ....\] we get
\[ \Rightarrow 2\left( e \right) + 2\left( {\dfrac{1}{{1!}} + \dfrac{2}{{2!}} + \dfrac{3}{{2!}} + .....} \right)\]
Now, if we see the second bracket can be written in the summation form as,
\[ \Rightarrow 2\left( e \right) + 2\sum\limits_{n = 1}^\infty {\dfrac{n}{{n!}}} \]
which is similar to the second term of the above given expression.
Therefore, on similarly solving and expanding it we get
\[ \Rightarrow 2\left( e \right) + 2\left( {1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + .....} \right)\]
\[ \Rightarrow 2\left( e \right) + 2\left( e \right) = 4e\]
On putting the value in the equation \[\left( 1 \right)\] we get,
\[4e + \left( e \right) + \left( {e - 1} \right)\]
\[ = 6e - 1\]
Thus, the value of \[\sum\limits_{n = 1}^\infty {\dfrac{{\left( {2{n^2} + n + 1} \right)}}{{n!}}} = 6e - 1\]
So, the correct answer is “Option 3”.

Note: While solving these types of problems, always try to simplify the terms in such a way that the expansion of any function becomes. Also, calculation plays an important role in these types of problems where we have to find the sum of the infinite series. In this problem one can go wrong while applying the expansion of the exponential function and the main thing to remember here is that the value of \[0! = 1\] not \[0\]
Some basic expansion of most commonly used functions are:
\[\sin z = z - \dfrac{{{z^3}}}{{3!}} + \dfrac{{{z^5}}}{{5!}} - \dfrac{{{z^7}}}{{7!}} + .....\]
\[\cos z = 1 - \dfrac{{{z^2}}}{{2!}} + \dfrac{{{z^4}}}{{4!}} - \dfrac{{{z^6}}}{{6!}} + .....\]
\[\sinh z = z + \dfrac{{{z^3}}}{{3!}} + \dfrac{{{z^5}}}{{5!}} + \dfrac{{{z^7}}}{{7!}} + .....\]
 \[\cosh z = 1 + \dfrac{{{z^2}}}{{2!}} + \dfrac{{{z^4}}}{{4!}} + \dfrac{{{z^6}}}{{6!}} + .....\]
\[{e^{ - z}} = 1 - z + \dfrac{{{z^2}}}{{2!}} - \dfrac{{{z^3}}}{{3!}} + .....\]
\[\ln \left( {1 + z} \right) = z - \dfrac{{{z^2}}}{2} + \dfrac{{{z^3}}}{3} - \dfrac{{{z^4}}}{4} + .....\]
\[\ln \left( {1 - z} \right) = - z - \dfrac{{{z^2}}}{2} - \dfrac{{{z^3}}}{3} - \dfrac{{{z^4}}}{4} - .....\]