Question:
The coefficient of $x^{-9}$ in the expansion of $\left(\dfrac{x^{2}}{2} - \dfrac{2}{x}\right)^{9}$ is}
The coefficient of $x^{-9}$ in the expansion of $\left(\dfrac{x^{2}}{2} - \dfrac{2}{x}\right)^{9}$ is}
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In binomial expansion, find the general term $T_{r+1}$, equate the power of $x$ to the required value, solve for $r$, then substitute back to get the coefficient.
Updated On: May 2, 2026
- 512
- $-512$
- 521
- 251
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Use the general term of the binomial expansion and equate the power of $x$ to $-9$.
Step 2: Detailed Explanation:
General term: \[T_{r+1} = \binom{9}{r}\left(\frac{x^2}{2}\right)^{9-r}\left(-\frac{2}{x}\right)^r = \binom{9}{r}(-1)^r\,2^{2r-9}\,x^{18-3r}\] Set $18 - 3r = -9 \Rightarrow r = 9$.
Coefficient $= \binom{9}{9}(-1)^9\,2^{18-9} = -2^9 = -512$.
Step 3: Final Answer:
The coefficient of $x^{-9}$ is $-512$.
Use the general term of the binomial expansion and equate the power of $x$ to $-9$.
Step 2: Detailed Explanation:
General term: \[T_{r+1} = \binom{9}{r}\left(\frac{x^2}{2}\right)^{9-r}\left(-\frac{2}{x}\right)^r = \binom{9}{r}(-1)^r\,2^{2r-9}\,x^{18-3r}\] Set $18 - 3r = -9 \Rightarrow r = 9$.
Coefficient $= \binom{9}{9}(-1)^9\,2^{18-9} = -2^9 = -512$.
Step 3: Final Answer:
The coefficient of $x^{-9}$ is $-512$.
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