Sum of real values of x for which middle term in binomial expansion of {left( {frac{{{x^3}}}{3

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7.Binomial Theorem

hard

The sum of the real values of $x$ for which the middle term in the binomial expansion of ${\left( {\frac{{{x^3}}}{3} + \frac{3}{x}} \right)^8}$ equals $5670$ is

A

$0$

B

$6$

C

$4$

D

$8$

(JEE MAIN-2019)

Solution

$5^{\text {th }}$ term will be the middle term.

$t_{4+1}=^{8} C_{4}\left(\frac{x^{3}}{3}\right)^{4}\left(\frac{3}{x}\right)^{4}=5670$

$=^{8} \mathrm{C}_{4} \cdot \mathrm{x}^{8}=5670$

$=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2} \mathrm{x}^{8}=5670$

$=x^{8}=\frac{567}{7}=81$

$=x^{8}-81=0$

$\Rightarrow$ Real value of $x=\pm \sqrt{3}$

Standard 11

Mathematics

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