Question

The sum of the real values of $x$ for which the middle term in the binomial expansion of ${{\left( \dfrac{{{x}^{3}}}{3}+\dfrac{3}{x} \right)}^{8}}$ equals 5670 is :
A. 6
B. 8
C. 0
D. 4

Answer 1

We know that in a binomial expansion of ${{\left( a+b \right)}^{n}}$, the middle term will be ${{\left( n+1 \right)}^{th}}$ term, where n = even. So, we will use this concept and the general term formula, ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$ to find the values and then we will equate it with the given value 5670 to get the answer.

Complete step-by-step answer:
In this question, the concept of binomial theorem is used on a specific side of the general and middle term concept. So, we need to understand those concepts as well. So, in an expansion of ${{\left( a+b \right)}^{n}}$, we have the expansion as ${}^{n}{{C}_{0}}{{a}^{0}}{{b}^{n}}+{}^{n}{{C}_{1}}{{a}^{1}}{{b}^{n-1}}+\ldots \ldots$. Now, if we consider the above expansion closely, we will get a general term for it as, ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{r}}{{b}^{n-r}}$. And we have that for the middle term, its dependent on n. So,
If n = even, then the middle term = ${{\left( \dfrac{n}{2}+1 \right)}^{th}}$ term.
And if n = odd, then there will be two middle terms, as, $MT1={{\left( \dfrac{n+1}{2} \right)}^{th}}$ term and $MT2={{\left( \dfrac{n+1}{2}+1 \right)}^{th}}$ term.
So, now, if we look at the given question, we have, ${{\left( \dfrac{{{x}^{3}}}{3}+\dfrac{3}{x} \right)}^{8}}$. So, we can say that in this we have the value of, $n=8,a=\dfrac{{{x}^{3}}}{3},b=\dfrac{3}{x}$. We have also been given the value of the middle term as 5670 and have been asked to find the value of $x$ such that the expansion of the middle term is 5670. So, we will first find the middle term, so we have the middle term as,
${{\left( \dfrac{8}{2}+1 \right)}^{th}}$ term $\Rightarrow {{5}^{th}}$ term.
Now, we will use the general formula and substitute the values of $a=\dfrac{{{x}^{3}}}{3},b=\dfrac{3}{x}$. So, we have,
${{T}_{r+1}}={}^{n}{{C}_{r}}{{\left( \dfrac{{{x}^{3}}}{3} \right)}^{r}}{{\left( \dfrac{3}{x} \right)}^{n-r}}\ldots \ldots \ldots \left( i \right)$
We have here, $r+1=5\Rightarrow r=4$. So, we will substitute the value of r in equation (i). So, we get,
$5670={}^{8}{{C}_{4}}{{\left( \dfrac{{{x}^{3}}}{3} \right)}^{4}}{{\left( \dfrac{3}{x} \right)}^{4}}\ldots \ldots \ldots \left( ii \right)$
Now, we know that, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, so putting the values, we get,

& {}^{8}{{C}_{4}}=\dfrac{8!}{4!4!} \\\ & \Rightarrow {}^{8}{{C}_{4}}=\dfrac{8\times 7\times 6\times 5\times 4!}{4!\times 4!}\text{ }as\left[ n!=n\left( n-1 \right)\left( n-2 \right)\ldots or\text{ }n!=n\left( n-1 \right)! \right] \\\ & \Rightarrow {}^{8}{{C}_{4}}=\dfrac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1} \\\ & \Rightarrow {}^{8}{{C}_{4}}=7\times 2\times 5=70 \\\ \end{aligned}$$ We will now substitute this value in equation (ii). So, we will get, $\begin{aligned} & 5670=70\times \dfrac{{{x}^{12}}}{{{3}^{4}}}\times \dfrac{{{3}^{4}}}{{{x}^{4}}} \\\ & \Rightarrow 5670=70\times {{x}^{8}} \\\ & \Rightarrow \dfrac{5670}{70}={{x}^{8}} \\\ & \Rightarrow {{x}^{8}}=81 \\\ & \Rightarrow x=\pm \sqrt{3} \\\ \end{aligned}$ Hence, the sum of the real values of $x=\sqrt{3}+\left( -\sqrt{3} \right)=0$. **So, the correct answer is “Option C”.** **Note:** We should know the use of the concept of general term and middle term of binomial expansion. We have to note the point in general terms that in ${{T}_{r+1}}$, $r+1$ exists as we know that the total number of terms in an expansion of ${{\left( a+b \right)}^{n}}$ is $n+1$. Sometimes the students don’t recollect the formula for general term and end up writing it as ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$ and go wrong. Do not try to expand the given term and then find the middle term, it will get very complicated as we have the power of 8 here.