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If the \(m\) th term is the middle term in expansion of \(\left(x^2-\frac{1}{2 x}\right)^{20}\). Find coefficient of \(T_{m+3}\).
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Verified Answer
The correct answer is:
\({ }^{20} \mathrm{C}_{13} 2^{-13}\)
Expansion of \(\left(x^2-\frac{1}{2 x}\right)^{20}\) contains ' 21 ' terms.
So middle term is, \(\frac{20}{2}+1=11\) term
\(\Rightarrow \quad m=11\)
Now,
\(\begin{aligned}
T_{m+3} & =T_{11+3}=T_{13+1} \\
& ={ }^{20} C_{13}\left(x^2\right)^{20-13}\left(-\frac{1}{2 x}\right)^{13} \\
& =\frac{{ }^{20} C_{13} x^7(-1)^{13}}{2^{13}, x^{13}}
\end{aligned}\)
So coefficient is, \({ }^{20} \mathrm{C}_{13} \cdot 2^{-13}{ }^{\prime \prime}\)
So middle term is, \(\frac{20}{2}+1=11\) term
\(\Rightarrow \quad m=11\)
Now,
\(\begin{aligned}
T_{m+3} & =T_{11+3}=T_{13+1} \\
& ={ }^{20} C_{13}\left(x^2\right)^{20-13}\left(-\frac{1}{2 x}\right)^{13} \\
& =\frac{{ }^{20} C_{13} x^7(-1)^{13}}{2^{13}, x^{13}}
\end{aligned}\)
So coefficient is, \({ }^{20} \mathrm{C}_{13} \cdot 2^{-13}{ }^{\prime \prime}\)
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