Question

The coefficient of ${x^3}$ in the expansion of ${3^x}$ is
A) $\dfrac{{\,{3^3}}}{6}$
B) $\dfrac{{{{\left( {\log 3} \right)}^3}}}{3}$
C) $\dfrac{3}{{3!}}$
D) None of the above

Answer 1

Hint: The very first and the most important hint of this question is that we have to use the expansion formula of ${a^x}$ to solve this question. From the expansion we have to find the coefficient of ${x^3}$. This question is just a simple application of the expansion formula.

Formula used: ${a^x} = 1 + \dfrac{{\left( {x\ln a} \right)}}{{1!}} + \dfrac{{{{\left( {x\ln a} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {x\ln a} \right)}^3}}}{{3!}} + - - - - - - $

Complete step by step answer:
In the above question, we have to find the value of coefficient of ${x^3}$ in the expansion of ${3^x}.$
For expansion, we know that
${a^x} = 1 + \dfrac{{\left( {x\ln a} \right)}}{{1!}} + \dfrac{{{{\left( {x\ln a} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {x\ln a} \right)}^3}}}{{3!}} + - - - - - - $
Now, put the value of $a = 3$ as required,
${3^x} = 1 + \dfrac{{\left( {x\ln 3} \right)}}{{1!}} + \dfrac{{{{\left( {x\ln 3} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {x\ln 3} \right)}^3}}}{{3!}} + - - - - - - $
On further calculation, we get
${3^x} = 1 + \dfrac{{x\left( {\ln 3} \right)}}{{1!}} + \dfrac{{{x^2}{{\left( {\ln 3} \right)}^2}}}{{2!}} + \dfrac{{{x^3}{{\left( {\ln 3} \right)}^3}}}{{3!}} + - - - - - - $
As we know that the coefficient of a particular value or variable is a term which is multiplied with it.
Therefore, we can see that the coefficient of ${x^3}$ in the above equation is $\dfrac{{{{\left( {\ln 3} \right)}^3}}}{{3!}}.$
Hence, the correct option is $\left( D \right)$.

Note: W can find the coefficient of any power of x. These formulas are known as binomial expansions. There are several other formulas of binomial expansion. We should also know the proof of these formulas which can be helpful in solving these types of problems.