The coefficient of {x^{-9}} in the expansion | vedclass

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Question

(B) The general term in the expansion of ${\left( \frac{x^2}{2} - \frac{2}{x} \right)^9}$ is given by ${T_{r+1}} = {}^9C_r \left( \frac{x^2}{2} \right

Vedclass · Accepted Answer

$-512$

Vedclass · Answer

(B) The general term in the expansion of ${\left( \frac{x^2}{2} - \frac{2}{x} \right)^9}$ is given by ${T_{r+1}} = {}^9C_r \left( \frac{x^2}{2} \right)^{9-r} \left( -\frac{2}{x} \right)^r$. Simplifying this expression,we get: ${T_{r+1}} = {}^9C_r \cdot \frac{x^{18-2r}}{2^{9-r}} \cdot \frac{(-2)^r}{x^r} = {}^9C_r \cdot \frac{(-2)^r}{2^{9-r}} \cdot x^{18-3r}$. To find the coefficient of ${x^{-9}}$,we set the exponent of $x$ equal to $-9$: $18 - 3r = -9$ $3r = 27$ $r = 9$. Substituting $r = 9$ into the coefficient part: Coefficient $= {}^9C_9 \cdot \frac{(-2)^9}{2^{9-9}} = 1 \cdot \frac{-512}{2^0} = -512$.

The coefficient of ${x^{-9}}$ in the expansion of ${\left( \frac{x^2}{2} - \frac{2}{x} \right)^9}$ is

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