JEE Mains · Maths · STD 11 - 7. binomial theoram
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 \(8\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\(5^{\text {th }}\) term will be the middle term. \(t_{4+1}=^{8} C_{4}\left(\frac{x^{3}}{3}\right)^{4}\left(\frac{3}{x}\right)^{4}=5670\) \(=^{8} \mathrm{C}_{4} \cdot \mathrm{x}^{8}=5670\) \(=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2} \mathrm{x}^{8}=5670\)…
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