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 $$\dfrac{(20)!}{(10)!(10)!} \,\cdot 2^{10}$$.Maths

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