KCET · Maths · Binomial Theorem
The ninth term of the expansion \(\left(3 x-\frac{1}{2 x}\right)^{8}\) is
- A \(\frac{1}{512 x^{9}}\)
- B \(\frac{-1}{512 x^{9}}\)
- C \(\frac{-1}{256 \cdot x^{8}}\)
- D \(\frac{1}{256 \cdot x^{8}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{256 \cdot x^{8}}\)
Step-by-step Solution
Detailed explanation
The general term of the expansion \((x+a)^{n}\) is \(T_{r+1}={ }^{n} C_{r} x^{n-r} a^{r} .\)
We have, \(\left(3 x-\frac{1}{2 x}\right)^{8}\).
Here, \(\quad r=8, x=3 x, a=\left(-\frac{1}{2 x}\right), n=8\)
\(\therefore\) Nineth term, \(T_{9}={ }^{8} C_{8}(3 x)^{8-8}\left(\frac{-1}{2 x}\right)^{8}\)
\[
=\frac{1}{256 \cdot x^{8}}
\]
We have, \(\left(3 x-\frac{1}{2 x}\right)^{8}\).
Here, \(\quad r=8, x=3 x, a=\left(-\frac{1}{2 x}\right), n=8\)
\(\therefore\) Nineth term, \(T_{9}={ }^{8} C_{8}(3 x)^{8-8}\left(\frac{-1}{2 x}\right)^{8}\)
\[
=\frac{1}{256 \cdot x^{8}}
\]
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