Identifying the Coefficient of ( x^7 ) in the Expansion of ( frac{4x}{5} - frac{5}{2x} )

In this article, we will explore the process of finding the coefficient of ( x^7 ) in the expansion of the given expression ( frac{4x}{5} - frac{5}{2x} ). This exercise involves a combination of algebraic manipulation and the application of the Binomial Theorem.

Introduction

Given the expression ( frac{4x}{5} - frac{5}{2x} ), it is crucial to understand that the term we are looking for, ( x^7 ), emerges from the expansion of this expression raised to an appropriate power. The problem can be approached by first simplifying the given expression and then applying the Binomial Theorem.

Simplification of the Expression

Let's begin by simplifying the given expression:

[ left( frac{4x}{5} - frac{5}{2x} right) ]

Since there is no exponents mentioned, we treat the expression as it is. Our goal is to identify the appropriate exponent ( n ) such that the expansion includes the term ( x^7 ).

Applying the Binomial Theorem

The General Expansion Theorem is a powerful tool in algebra that helps in expanding expressions of the form ( (a b)^n ). Here, ( a frac{4x}{5} ) and ( b -frac{5}{2x} ).

Using the Binomial Theorem, the expansion is given by:

[ left( frac{4x}{5} - frac{5}{2x} right)^n sumlimits_{k0}^{n} binom{n}{k} left( frac{4x}{5} right)^{n-k} left( -frac{5}{2x} right)^k ]

Let's break this down.

[ left( frac{4x}{5} right)^{n-k} left( -frac{5}{2x} right)^k left( frac{4x}{5} right)^{n-k} left( -1 right)^k left( frac{5}{2x} right)^k ]

Combining these terms, we get:

[ left( frac{4x}{5} right)^{n-k} left( -1 right)^k left( frac{5}{2x} right)^k frac{4^{n-k} x^{n-k}}{5^{n-k}} cdot (-1)^k cdot frac{5^k}{2^k x^k} ]

[ (-1)^k cdot frac{4^{n-k} cdot 5^k}{5^{n-k} cdot 2^k} cdot x^{n-2k} ]

[ (-1)^k cdot frac{2^{2(n-k)} cdot 5^k}{5^{n-k} cdot 2^k} cdot x^{n-2k} ]

[ (-1)^k cdot 2^{2(n-k)-k} cdot 5^{k-n} cdot x^{n-2k} ]

Finding the Coefficient of ( x^7 )

To find the term involving ( x^7 ), we set the exponent of ( x ) to 7:

[ n - 2k 7 ]

Solving for ( k ) in terms of ( n ):

[ n - 2k 7 ] [ 2k n - 7 ] [ k frac{n - 7}{2} ]

For ( k ) to be an integer, ( n - 7 ) must be even, implying ( n ) must be an odd number greater than or equal to 7.

Given ( k frac{n - 7}{2} ), the coefficient of ( x^7 ) is:

[ (-1)^k cdot 2^{2(n-k) - k} cdot 5^{k-n} ]

Substituting ( k frac{n - 7}{2} ) into the expression for the coefficient:

[ (-1)^{frac{n - 7}{2}} cdot 2^{2(n - frac{n - 7}{2}) - frac{n - 7}{2}} cdot 5^{frac{n - 7}{2} - n} ]

[ (-1)^{frac{n - 7}{2}} cdot 2^left(2n - n 7 - frac{n - 7}{2}right) cdot 5^{-frac{n - 7}{2}} ]

[ (-1)^{frac{n - 7}{2}} cdot 2^left(n 7 - frac{n - 7}{2}right) cdot 5^{-frac{n - 7}{2}} ]

[ (-1)^{frac{n - 7}{2}} cdot 2^left(frac{2n 14 - n 7}{2}right) cdot 5^{-frac{n - 7}{2}} ]

[ (-1)^{frac{n - 7}{2}} cdot 2^left(frac{n 21}{2}right) cdot 5^{-frac{n - 7}{2}} ]

Conclusion

In conclusion, to find the coefficient of ( x^7 ) in the expansion of the given expression, the exponent ( n ) must be an odd number greater than 7. The coefficient is determined by the values of ( n ) and ( k frac{n - 7}{2} ).

These steps highlight the power of algebraic manipulation and the Binomial Theorem in solving complex polynomial expansions. This problem and its solution are not only intellectually stimulating but also valuable in understanding the intricacies of polynomial expansions.