Middle term in expansion of \([(2 / x)-3 x y]^{12}\) is (a) \(14370048 \mathrm{y}^{6}\) (b) \(14370024 \mathrm{y}^{6}\) (c) \(43110144 \mathrm{y}^{6}\) (d) \(43110124 \mathrm{y}^{6}\)

Short Answer

Expert verified
The middle term in the expansion of \([(2/x)-3xy]^{12}\) is (a) \(14370048 \mathrm{y}^{6}\).

Step by step solution

01

Find the middle term's index

Since n is even (12 is even), the middle term will be at position \(\frac{n}{2} + 1 = \frac{12}{2} + 1 = 7\). So, we are looking for the 7th term in the expansion: \(T_7\).
02

Apply the Binomial theorem to find the middle term

Using the Binomial theorem, we can write the 7th term as \[T_7 = ^{12}C_6(2/x)^{12-6}(-3xy)^6\]
03

Calculate the binomial coefficient

Now, let's calculate the binomial coefficient: \[^{12}C_6 = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!} = 924\]
04

Simplify the term

Now we plug in the binomial coefficient and simplify the term: \[T_7 = 924\left(\frac{2}{x}\right)^6(-3xy)^6\] \[T_7 = 924\left(\frac{2^6}{x^6}\right)(-3^6x^6y^6)\] \[T_7 = 924(64)(-729x^6y^6)\] \[T_7 = 14370048y^6\] So, the middle term in the expansion of \([(2/x)-3xy]^{12}\) is \(14370048y^6\). The answer is (a) \(14370048 \mathrm{y}^{6}\).

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