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$6^{\text {th }}$ term in expansion of $\left(2 x^2-\frac{1}{3 x^2}\right)^{10}$
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Verified Answer
The correct answer is:
$-\frac{896}{27}$
Applying $T_{r+1}={ }^n C_r x^{n-r} a^r$ for $(x+a)^n$
Hence $T_6={ }^{10} C_5\left(2 x^2\right)^5\left(-\frac{1}{3 x^2}\right)^5$
$=-\frac{10!}{5!5!} 32 \times \frac{1}{243}=-\frac{896}{27}$
Hence $T_6={ }^{10} C_5\left(2 x^2\right)^5\left(-\frac{1}{3 x^2}\right)^5$
$=-\frac{10!}{5!5!} 32 \times \frac{1}{243}=-\frac{896}{27}$
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