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The middle term in the expansion of $\left(4 x^3-\frac{15}{4 x}\right)^8$ is
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Verified Answer
The correct answer is:
$70\left(15 x^2\right)^4$
Here, $n=8$ (even)
$\therefore\left(\frac{n}{2}+1\right)$ th term will be middle term.
So,
$\begin{aligned}
T_{4+1} & ={ }^8 C_4 \cdot\left(4 x^3\right)^4 \cdot\left(\frac{-15}{4 x}\right)^4 \\
& =\frac{8 !}{4 ! \cdot 4 !} \cdot 256 \times x^{12} \cdot \frac{225 \times 225}{256 \times x^4} \\
& =\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \cdot(15)^4 \cdot\left(x^2\right)^4 \\
& =70\left(15 x^2\right)^4
\end{aligned}$
$\therefore\left(\frac{n}{2}+1\right)$ th term will be middle term.
So,
$\begin{aligned}
T_{4+1} & ={ }^8 C_4 \cdot\left(4 x^3\right)^4 \cdot\left(\frac{-15}{4 x}\right)^4 \\
& =\frac{8 !}{4 ! \cdot 4 !} \cdot 256 \times x^{12} \cdot \frac{225 \times 225}{256 \times x^4} \\
& =\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \cdot(15)^4 \cdot\left(x^2\right)^4 \\
& =70\left(15 x^2\right)^4
\end{aligned}$
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