Find the coefficient of ${x^n}$ in the expansion of ${e^{a - bx}}$.
Answer
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Hint: ${e^{a - bx}}$ can be written as ${e^a}.{e^{ - bx}}$. And then the expansion of ${e^x}$ is $1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + ....$
Use this expansion for ${e^{ - bx}}$ and find the general term when it is multiplied by ${e^a}$.
Complete step-by-step answer:
According to the question, we have to find out the coefficient of ${x^n}$ in the expansion of ${e^{a - bx}}$.
We can write ${e^{a - bx}}$ as:
$ \Rightarrow {e^{a - bx}} = {e^a}.{e^{ - bx}} .....(i)$
We know the expansion of ${e^x}$ is:
$ \Rightarrow {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + ....$
Using this expansion for ${e^{ - bx}}$ in equation $(i)$, we’ll get:
\[ \Rightarrow {e^{a - bx}} = {e^a} \times \left[ {1 + \dfrac{{\left( { - bx} \right)}}{{1!}} + \dfrac{{{{\left( { - bx} \right)}^2}}}{{2!}} + \dfrac{{{{\left( { - bx} \right)}^3}}}{{3!}}...\dfrac{{{{\left( { - bx} \right)}^n}}}{{n!}}...} \right]\]
In the above expansion, ${x^n}$ will occur for the term ${e^a}.\dfrac{{{{\left( { - bx} \right)}^n}}}{{n!}}$.
This term can be written as ${e^a}.\dfrac{{{{\left( { - b} \right)}^n}}}{{n!}}{x^n}$.
Thus, the coefficient of ${x^n}$ in the expansion of ${e^{a - bx}}$ is ${e^a}.\dfrac{{{{\left( { - b} \right)}^n}}}{{n!}}$.
Note: We could have used the expansion of ${e^{a - bx}}$ directly as:
$ \Rightarrow {e^{a - bx}} = 1 + \dfrac{{\left( {a - bx} \right)}}{{1!}} + \dfrac{{{{\left( {a - bx} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {a - bx} \right)}^3}}}{{3!}} + ....$
Although theoretically we will get the same result as above but it’s not possible to find it out in this case because every term will contain a mixture of different powers of $x$ and the expansion is also going up to infinity.
Use this expansion for ${e^{ - bx}}$ and find the general term when it is multiplied by ${e^a}$.
Complete step-by-step answer:
According to the question, we have to find out the coefficient of ${x^n}$ in the expansion of ${e^{a - bx}}$.
We can write ${e^{a - bx}}$ as:
$ \Rightarrow {e^{a - bx}} = {e^a}.{e^{ - bx}} .....(i)$
We know the expansion of ${e^x}$ is:
$ \Rightarrow {e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + ....$
Using this expansion for ${e^{ - bx}}$ in equation $(i)$, we’ll get:
\[ \Rightarrow {e^{a - bx}} = {e^a} \times \left[ {1 + \dfrac{{\left( { - bx} \right)}}{{1!}} + \dfrac{{{{\left( { - bx} \right)}^2}}}{{2!}} + \dfrac{{{{\left( { - bx} \right)}^3}}}{{3!}}...\dfrac{{{{\left( { - bx} \right)}^n}}}{{n!}}...} \right]\]
In the above expansion, ${x^n}$ will occur for the term ${e^a}.\dfrac{{{{\left( { - bx} \right)}^n}}}{{n!}}$.
This term can be written as ${e^a}.\dfrac{{{{\left( { - b} \right)}^n}}}{{n!}}{x^n}$.
Thus, the coefficient of ${x^n}$ in the expansion of ${e^{a - bx}}$ is ${e^a}.\dfrac{{{{\left( { - b} \right)}^n}}}{{n!}}$.
Note: We could have used the expansion of ${e^{a - bx}}$ directly as:
$ \Rightarrow {e^{a - bx}} = 1 + \dfrac{{\left( {a - bx} \right)}}{{1!}} + \dfrac{{{{\left( {a - bx} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {a - bx} \right)}^3}}}{{3!}} + ....$
Although theoretically we will get the same result as above but it’s not possible to find it out in this case because every term will contain a mixture of different powers of $x$ and the expansion is also going up to infinity.
Last updated date: 24th Sep 2023
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