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Chapter 7: Problem 494

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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Most popular questions from this chapter

\((10.1)^{5}=\) (a) \(105101.501\) (b) \(105101.0501\) (c) \(105101.00501\) (d) \(105101.05001\)

Co-efficient of \(\mathrm{x}^{53}\) in \(\sqrt{ }^{100} \sum_{\mathrm{r}=0}{ }^{100} \mathrm{C}_{\mathrm{r}}(\mathrm{x}-5)^{100-\mathrm{r}} 4^{\mathrm{r}}\) is (a) \({ }^{100} \mathrm{C}_{53}\) (b) \({ }^{100} \mathrm{C}_{48}\) (c) \(-{ }^{100} \mathrm{C}_{53}\) (d) \({ }^{100} \mathrm{C}_{51}\)

Coefficients of middle terms in expansion of \(\left[2-\left(\mathrm{x}^{3} / 3\right)\right]^{7}\) are... (a) \(-[(560) /(27)],-[(280) /(81)]\) (b) \([(560) /(27)],-[(280) /(81)]\) (c) \(-[(560) /(27)],[(280) /(81)]\) (d) \([(560) /(27)],[(280) /(81)]\)

\({ }^{\mathrm{n}} \sum_{\mathrm{r}=0}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} 4^{\mathrm{r}}=\) (a) \(4^{\mathrm{n}}\) (b) \(5^{\mathrm{n}}\) (c) \(4^{-n}\) (d) \(5^{-n}\)

If coefficient of \(2^{\text {nd }}, 3^{\text {rd }}\) and \(4^{\text {th }}\) terms re in A. P for \((1+x)^{n}\) then \(\mathrm{n}=\) (a) 28 (b) 14 (c) 7 (d) \((7 / 2)\)
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