Question

The coefficient of \[{x^n}\] in the expansion of \[{e^{{e^x}}}\] is
A. \[{x^n}\dfrac{1}{{\left( {n + 1} \right)!}}\left( {\dfrac{{{1^n}}}{{1!}} + \dfrac{{{2^n}}}{{2!}} + \dfrac{{{3^n}}}{{3!}} + ...\infty } \right)\]
B. \[{x^n}\dfrac{1}{{n!}}\left( {\dfrac{{{1^n}}}{{1!}} + \dfrac{{{2^n}}}{{2!}} + \dfrac{{{3^n}}}{{3!}} + ...\infty } \right)\]
C. \[\dfrac{1}{{n!}}\left( {\dfrac{{{1^n}}}{{1!}} + \dfrac{{{2^n}}}{{2!}} + \dfrac{{{3^n}}}{{3!}} + ...\infty } \right)\]
D. \[\dfrac{1}{{n!}}\left( {\dfrac{{{1^n}}}{{1!}} + \dfrac{{{3^n}}}{{3!}} + \dfrac{{{5^n}}}{{5!}} + ...\infty } \right)\]

Answer 1

- Hint: The given function is an exponential function. So, write the series expansion for the exponential function \[{e^x}\]. Substitute the expansion of \[{e^x}\] in the given function \[{e^{{e^x}}}\]. The \[\left( {n + 1} \right)\]-th term of the series contains \[{x^n}\]. So, the coefficient of \[{x^n}\] will be found from the \[\left( {n + 1} \right)\]-th term.

Formula Used:
The series expansion for \[{e^x}\] is \[1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .....\infty \]

Complete step-by-step answer:
The given function is an exponential function.
We know the series expansion for the exponential function \[{e^x}\], which is \[1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .....\infty \]
So, \[{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ..... + \dfrac{{{x^n}}}{{n!}} + .....\infty - - - - - \left( i \right)\]
But there is \[{e^x}\] at the place of \[x\] in the given function.
So, if we replace \[x\] by \[{e^x}\], we get
\[{e^{{e^x}}} = 1 + \dfrac{{{e^x}}}{{1!}} + \dfrac{{{{\left( {{e^x}} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {{e^x}} \right)}^3}}}{{3!}} + ..... + \dfrac{{{{\left( {{e^x}} \right)}^n}}}{{n!}} + .....\infty - - - - - \left( {ii} \right)\]
Using the formula \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], we can write \[{\left( {{e^x}} \right)^2} = {e^{2x}}\] and \[{\left( {{e^x}} \right)^3} = {e^{3x}}\]
From equation \[\left( {ii} \right)\], we get
\[{e^{{e^x}}} = 1 + \dfrac{{{e^x}}}{{1!}} + \dfrac{{{e^{2x}}}}{{2!}} + \dfrac{{{e^{3x}}}}{{3!}} + ..... + \dfrac{{{e^{nx}}}}{{n!}} + .....\infty - - - - - \left( {iii} \right)\]
Obtain the expansions of \[{e^{2x}}\] and \[{e^{3x}}\] using the expansion given by \[\left( i \right)\]
Replacing \[x\] by \[2x\] in \[\left( i \right)\], we get
\[{e^{2x}} = 1 + \dfrac{{\left( {2x} \right)}}{{1!}} + \dfrac{{{{\left( {2x} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {2x} \right)}^3}}}{{3!}} + ..... + \dfrac{{{{\left( {2x} \right)}^n}}}{{n!}} + .....\infty - - - - - \left( {iv} \right)\]
Replacing \[x\] by \[3x\] in \[\left( i \right)\], we get
\[{e^{3x}} = 1 + \dfrac{{\left( {3x} \right)}}{{1!}} + \dfrac{{{{\left( {3x} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {3x} \right)}^3}}}{{3!}} + ..... + \dfrac{{{{\left( {3x} \right)}^n}}}{{n!}} + .....\infty - - - - - \left( v \right)\]
Substituting the expansions of \[{e^x}\], \[{e^{2x}}\] and \[{e^{3x}}\] given by \[\left( i \right)\], \[\left( {iv} \right)\] and \[\left( v \right)\] in the expansion of \[{e^{{e^x}}}\] given by \[\left( {iii} \right)\], we get
\[{e^{{e^x}}} = 1 + \dfrac{1}{{1!}}\left( {1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ..... + \dfrac{{{x^n}}}{{n!}} + .....\infty } \right) + \dfrac{1}{{2!}}\left( {1 + \dfrac{{\left( {2x} \right)}}{{1!}} + \dfrac{{{{\left( {2x} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {2x} \right)}^3}}}{{3!}} + ..... + \dfrac{{{{\left( {2x} \right)}^n}}}{{n!}} + .....\infty } \right) + \dfrac{1}{{3!}}\left( {1 + \dfrac{{\left( {3x} \right)}}{{1!}} + \dfrac{{{{\left( {3x} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {3x} \right)}^3}}}{{3!}} + ..... + \dfrac{{{{\left( {3x} \right)}^n}}}{{n!}} + .....\infty } \right) + .....\infty - - - - - \left( {vi} \right)\]
It is required to find out the coefficient of \[{x^n}\] in the expansion of the given function \[{e^{{e^x}}}\], which is given in \[\left( {vi} \right)\]
The term \[{x^n}\] is present in \[\left( {n + 1} \right)\]-th term of each bracket in the right-hand side of the expansion given by \[\left( {vi} \right)\]
If we arrange the terms having \[{x^n}\], we get the term \[\dfrac{1}{{n!}}\left( {\dfrac{{{1^n}}}{{1!}} + \dfrac{{{2^n}}}{{2!}} + \dfrac{{{3^n}}}{{3!}} + .....} \right){x^n}\]
So, the coefficient of \[{x^n}\] in the given series is \[\dfrac{1}{{n!}}\left( {\dfrac{{{1^n}}}{{1!}} + \dfrac{{{2^n}}}{{2!}} + \dfrac{{{3^n}}}{{3!}} + .....} \right)\]
Hence, option C is correct.

Note:- In the series expansion of \[{e^x}\], all the terms are positive for positive values of \[x\] but for the negative values of \[x\], the signs of the terms alter. The denominators consist of the factorial of the natural numbers.