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Chapter 7: Problem 493

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) \([(560) /(27)],[(280) /(81)]\)

Short Answer

Expert verified
The coefficients of the middle terms in the expansion of \(\left[2-\left(\mathrm{x}^{3} / 3\right)\right]^{7}\) are \(\frac{560}{27}\) and \(-\frac{280}{81}\) (Option b).

Step by step solution

01

Identify the Binomial Expansion

We are given the binomial expansion \(\left[2-\left(\mathrm{x}^{3} / 3\right)\right]^{7}\).
02

Determine the Middle Terms

Since there are 7 terms in the expansion (from the exponent), there are two middle terms. These will be the 4th and 5th terms in the expansion.
03

Use the Binomial Coefficient Formula

We will use the binomial coefficient formula to compute the coefficients of the 4th and 5th terms. The formula is: \(_{n} C_{r} = \frac{n!}{r! * (n-r)!}\) Where n is the total number of terms, r is the term in question, and ! represents the factorial of a given number.
04

Compute the Coefficients of the 4th Term

We want to find the coefficient of the 4th term, so we need to calculate \( _{7} C_{3} \): \(_{7} C_{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3! * 4!} = \frac{7*6*5}{3*2*1} = \frac{210}{6} = 35\) The 4th term, including the coefficient, is \((35) \cdot (2)^{4}\cdot\left(-\frac{x^3}{3}\right)^{3}\). Simplify this expression: \(35 \cdot 16 \cdot \left(-\frac{x^9}{27}\right) = \displaystyle\frac{-560x^9}{27}\) Therefore, the coefficient of the 4th term is \(\displaystyle\frac{-560}{27}\).
05

Compute the Coefficients of the 5th Term

We want to find the coefficient of the 5th term, so we need to calculate \( _{7} C_{4} \): \(_{7} C_{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! * 3!} = \frac{7*6*5}{4*3*2} = \frac{210}{24} = 35\cdot2 = 70\) The 5th term, including the coefficient, is \((70) \cdot (2)^{3}\cdot\left(-\frac{x^3}{3}\right)^{4}\). Simplify this expression: \(70 \cdot 8 \cdot \left(\frac{x^{12}}{81}\right) = \displaystyle\frac{560x^{12}}{81}\) Therefore, the coefficient of the 5th term is \(\displaystyle\frac{560}{81}\).
06

Compare the Coefficients with the Given Options

We found the coefficients to be: \(\displaystyle\frac{-560}{27}\) and \(\displaystyle\frac{560}{81}\). Comparing these with the given options, we can see that the correct answer is: (b) \([(\frac{560}{27})],[-(\frac{280}{81})]\)

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Most popular questions from this chapter

\(\left|\begin{array}{c}\mathrm{n}-1 \\\ 1\end{array}\right|+\left|\begin{array}{c}\mathrm{n}-1 \\\ 2\end{array}\right|+\ldots+\left|\begin{array}{l}\mathrm{n}-1 \\\ \mathrm{n}-1\end{array}\right|=\underline{ } ; \mathrm{n}>1\) (a) \(2^{\mathrm{n}}-1\) (b) \(2^{\mathrm{n}-2}\) (c) \(2^{\mathrm{n}-1}-1\) (d) \(2^{\mathrm{n}-1}\)

It coefficients of middle terms in expansion of \((1+\lambda \mathrm{x})^{8}\) and \((1-\lambda \mathrm{x})^{6}\) are equal then \(\lambda=\) (a) \((2 / 7)\) (b) \([(-2) / 7]\) (c) \([(-3) / 7]\) (d) None of these

If middle term is \(\mathrm{kx}^{\mathrm{m}}\) in expansion of \([\mathrm{x}+(1 / \mathrm{x})]^{12}\) then \(\mathrm{m}=\) (a) \(-2\) (b) \(-1\) (c) 0 (d) 1

\(\left[(\sqrt{2}+1)^{8}\right]=\ldots \ldots \ldots ;\) where [] is integer part function. (a) 1151 (b) 1152 (c) 1153 (a) 1154

Index number of middle term in expansion of \(\left[1+a+\left(a^{2} / 4\right)\right]^{n}\) is (a) \((\mathrm{n} / 2)+1\) (b) \([(\mathrm{n}+2) / 2]\) (c) \(n+1\) (d) \([(\mathrm{n}+3) / 2]\)
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