If n is even and the middle term in the expan | 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-r}

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The general term in the expansion of $\left(x^2+\frac{1}{x}ight)^n$ is given by $T_{r+1} = {}^nC_r (x^2)^{n-r} (x^{-1})^r = {}^nC_r x^{2n-2r-r} = {}^nC_r x^{2n-3r}$.Since $n$ is even,the middle term is the $(\frac{n}{2} + 1)$-th term,where $r = \frac{n}{2}$.Substituting $r = \frac{n}{2}$ into the general term expression:$T_{\frac{n}{2}+1} = {}^nC_{\frac{n}{2}} x^{2n-3(\frac{n}{2})} = {}^nC_{\frac{n}{2}} x^{\frac{n}{2}}$.Given that the middle term is $924 x^6$,we have $\frac{n}{2} = 6$,which implies $n = 12$.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,$n = 12$ is the correct value.

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

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