In the expansion of (x^2 - 2x)^{10},the coeff | vedclass

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(C) The general term in the expansion of $(x^2 - 2x)^{10}$ is given by $T_{r+1} = \binom{10}{r} (x^2)^{10-r} (-2x)^r$.This simplifies to $T_{r+1} = \b

Vedclass · Accepted Answer

$3360$

Vedclass · Answer

(C) The general term in the expansion of $(x^2 - 2x)^{10}$ is given by $T_{r+1} = \binom{10}{r} (x^2)^{10-r} (-2x)^r$.This simplifies to $T_{r+1} = \binom{10}{r} x^{20-2r} (-2)^r x^r = \binom{10}{r} (-2)^r x^{20-r}$.We want the coefficient of $x^{16}$,so we set the exponent $20-r = 16$,which gives $r = 4$.Substituting $r = 4$ into the expression,the coefficient is $\binom{10}{4} (-2)^4$.Calculating the value: $\binom{10}{4} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$.$(-2)^4 = 16$.Thus,the coefficient is $210 	imes 16 = 3360$.

In the expansion of $(x^2 - 2x)^{10}$,the coefficient of $x^{16}$ is

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