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Question

Find the coefficient of $$x^{10}$$ and $$x^9$$ in the expansion of $$\left(2x^2 \, - \, \dfrac{1}{x} \right)^{20}$$


Solution

$$T_{r \, + \, 1} \, = \, ^{20}C_r \, \, (2x^2)^{20 \, - \, r} \, \, (- 1/x)^r$$
$$= \, (-1)^r \, \, ^{20}C_r \, \, 2^{20 \, - \, r} \, \, x^{40\, - \, 2r \, - \, r}$$
$$\therefore  \, 40 \, -\, 3r \, = \, 10$$          or        9
$$\therefore  \, 3r \, = \, 30$$       or     31.
$$\therefore$$ $$r \, = \, 10$$ the other value does not give integral value of r so that there will be term of $$x^9$$ .
Putting $$r \, = \, 10$$
$$T_{11} \, = \, (-1)^{10} \, \, ^{20}C_{10} \, 2^{20 \, - \, 10} \, x^{40 \, - \, 30}$$
$$= \, \dfrac{(20)!}{(10)!(10)!}  \, 2^{10} \, \, x^{10}$$
Hence the required coefficient is $$\dfrac{(20)!}{(10)!(10)!}  \,\cdot  2^{10}$$.

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