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What is the middle term in the expansion of $\left(1-\frac{x}{2}\right)^{8}$ ?
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Verified Answer
The correct answer is:
$\frac{35 x^{4}}{8}$
Since $n=8$ is even number therefore middle term
$=\left(\frac{n}{2}+1\right)^{\mathrm{th}}$ term $=(4+1)=5^{\mathrm{th}}$ term
Hence, $T_{5}={ }^{8} C_{4}(1)^{4}\left(-\frac{x}{2}\right)^{4}$
$=\frac{8 !}{4 ! 4 !} \times \frac{x^{4}}{16}=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \cdot \frac{x^{4}}{16}=\frac{70 x^{4}}{16}=\frac{35 x^{4}}{8}$
$=\left(\frac{n}{2}+1\right)^{\mathrm{th}}$ term $=(4+1)=5^{\mathrm{th}}$ term
Hence, $T_{5}={ }^{8} C_{4}(1)^{4}\left(-\frac{x}{2}\right)^{4}$
$=\frac{8 !}{4 ! 4 !} \times \frac{x^{4}}{16}=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \cdot \frac{x^{4}}{16}=\frac{70 x^{4}}{16}=\frac{35 x^{4}}{8}$
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