The coefficient of x^2 in the expansion of (2 | vedclass

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(B) The general term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.For the expansion $(2x^2 + x^{-1})^{10}$,

Vedclass · Accepted Answer

$3360$

Vedclass · Answer

(B) The general term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.For the expansion $(2x^2 + x^{-1})^{10}$,the general term is $T_{r+1} = \binom{10}{r} (2x^2)^{10-r} (x^{-1})^r$.Simplifying this,we get $T_{r+1} = \binom{10}{r} 2^{10-r} x^{20-2r} x^{-r} = \binom{10}{r} 2^{10-r} x^{20-3r}$.To find the coefficient of $x^2$,we set the exponent of $x$ equal to $2$:$20 - 3r = 2$$3r = 18$$r = 6$.Substituting $r = 6$ into the expression for the coefficient:Coefficient $= \binom{10}{6} 2^{10-6} = \binom{10}{4} 2^4$.Calculating the value: $\binom{10}{4} = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$.Coefficient $= 210 	imes 16 = 3360$.

The coefficient of $x^2$ in the expansion of $(2x^2 + \frac{1}{x})^{10}, x \neq 0$,is:

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