The middle term in the expansion of left(4x^3 | vedclass

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Question

(C) Given the binomial expansion $\left(4x^3 - \frac{15}{4x}\right)^8$,where $n = 8$.Since $n$ is even,the middle term is the $\left(\frac{n}{2} + 1\r

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$70(15x^2)^4$

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(C) Given the binomial expansion $\left(4x^3 - \frac{15}{4x}ight)^8$,where $n = 8$.Since $n$ is even,the middle term is the $\left(\frac{n}{2} + 1ight)$-th term,which is the $5$-th term $(T_5)$.Using the general term formula $T_{r+1} = {}^nC_r \cdot a^{n-r} \cdot b^r$:$T_{4+1} = {}^8C_4 \cdot (4x^3)^4 \cdot \left(-\frac{15}{4x}ight)^4$$T_5 = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} \cdot (4^4 \cdot x^{12}) \cdot \left(\frac{15^4}{4^4 \cdot x^4}ight)$$T_5 = 70 \cdot x^{12-4} \cdot 15^4$$T_5 = 70 \cdot (15x^2)^4$

The middle term in the expansion of $\left(4x^3 - \frac{15}{4x}\right)^8$ is

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