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Chapter 11: Problem 56

Find the middle term in the expansion of \(\left(\frac{1}{x}-x^{2}\right)^{12}\)

Short Answer

Expert verified
The middle terms of the expansion \(\left(\frac{1}{x}-x^{2}\right)^{12}\) are \( -792x^{15} \) and \( 924x^{12} \).

Step by step solution

01

Identify the values

Identify the values for the binomial theorem formula. Here, n = 12, a = \( \frac{1}{x} \) and b = \( -x^2 \)
02

Practical implementation of Binomial Theorem

In a binomial expansion, there are (n+1) terms. Hence, in this case, there are 13 terms, which results in two middle terms. The 6th term and the 7th term are the middle terms of the expansion. They are calculated using the formula for Binomial theorem.
03

Calculate the 6th term

Use the formula to find the 6th term: \( T_{6} = C_{12}^{5} * \left(\frac{1}{x}\right)^{7} * (-x^{2})^{5} = 792*(-x^{3})^5 = -792x^{15} \)
04

Calculate the 7th term

Now use the formula to find the 7th term: \( T_{7} = C_{12}^{6} * \left(\frac{1}{x}\right)^{6} * (-x^{2})^{6} = 924*x^{12} = 924x^{12} \)

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