If the middle term in the expansion of {left( | vedclass

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

Vedclass · Accepted Answer

$12$

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(B) The general term in the expansion of ${\left( {x^2 + \frac{1}{x}} ight)^n}$ is given by $T_{r+1} = ^nC_r (x^2)^{n-r} (x^{-1})^r = ^nC_r x^{2n-2r} x^{-r} = ^nC_r x^{2n-3r}$.Since the middle term is given,$n$ must be an even integer. The middle term is the $\left( \frac{n}{2} + 1 ight)^{th}$ term,so $r = \frac{n}{2}$.Substituting $r = \frac{n}{2}$ into the general term expression,we get $T_{\frac{n}{2}+1} = ^nC_{n/2} x^{2n - 3(n/2)} = ^nC_{n/2} x^{n/2}$.Given that the middle term is $924x^6$,we equate the powers of $x$: $\frac{n}{2} = 6 \Rightarrow n = 12$.Checking the coefficient: $^{12}C_6 = \frac{12 	imes 11 	imes 10 	imes 9 	imes 8 	imes 7}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = 924$.Thus,$n = 12$.

If the middle term in the expansion of ${\left( {x^2 + \frac{1}{x}} \right)^n}$ is $924x^6$,then $n = $

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