Problem 8 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Find the 7 th term in the expansion \(f\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{9}\)

Step-by-Step Solution

Answer

The 7th term is \(\frac{10500}{x^3}\).

Step 1: Understand the Binomial Expansion

The problem is asking for a specific term in the expansion of \( \left(\frac{4x}{5} - \frac{5}{2x}\right)^9 \). This is a binomial expression which can be expanded using the binomial theorem.

Step 2: Write the Binomial Theorem Formula

The binomial theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In this case, \(a = \frac{4x}{5}\), \(b = -\frac{5}{2x}\), and \(n = 9\).

Step 3: Identify the 7th Term

The \((k+1)\)th term in the expansion is given by \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\). For the 7th term, \(k = 6\).

Step 4: Calculate \(\binom{9}{6}\)

To find \(\binom{9}{6}\), use the combination formula: \(\binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\).

Step 5: Compute \(a^{n-k}\) and \(b^k\)

For \(a^{n-k}\), compute \(\left(\frac{4x}{5}\right)^{9-6} = \left(\frac{4x}{5}\right)^3 = \frac{64x^3}{125}\). For \(b^{k}\), compute \(\left(-\frac{5}{2x}\right)^6 = \frac{15625}{64x^6}\).

Step 6: Multiply Components Together for \(T_7\)

Multiply the components to find the 7th term: \[ T_7 = 84 \times \frac{64x^3}{125} \times \frac{15625}{64x^6} = 84 \times \frac{15625}{125x^3} = 84 \times \frac{125}{x^3}\].

Step 7: Simplify \(T_7\)

After the calculation, simplify to find: \( T_7 = 84 \times 125/x^3 = 10500/x^3\).

Key Concepts

Binomial Theorem

The Binomial Theorem is an essential tool in algebra that allows us to expand expressions in the form of \((a + b)^n\). It provides a structured way to write the expanded form without directly multiplying everything out, which can be tedious. The theorem itself is represented as:

• \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).

This formula tells us how to find each term in the expanded expression for a given \(n\). In our exercise, the expression \(\left(\frac{4x}{5} - \frac{5}{2x}\right)^9\) can be expanded using this theorem, with \(a = \frac{4x}{5}\), \(b = -\frac{5}{2x}\), and \(n = 9\). By recognizing that this is a binomial expression, we can apply the theorem to find any specific term we are interested in, such as the 7th term in this case.

Combination Formula

The Combination Formula is utilized to determine the number of ways to choose \(k\) elements from a set of \(n\) elements, which is crucial in finding terms of a binomial expansion. The formula is given by:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

This tells us the coefficient needed for each term in the expansion. In practice, calculating a binomial coefficient like \(\binom{9}{6}\) involves evaluating:

• \(\frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\)

Knowing how to compute these coefficients is key to accessing each possible term in the binomial expansion process.

Exponents

Understanding how to manage exponents is crucial while working through a binomial expansion. Each term in the expansion is defined partly through powers of \(a\) and \(b\). Take, for instance, the term with a power, expressed as \(a^{n-k} b^k\).For example:

• To find \(a^{n-k}\), you compute \(\left(\frac{4x}{5}\right)^{9-6} = \frac{64x^3}{125}\).
• As for \(b^k\), it becomes \(\left(-\frac{5}{2x}\right)^6 = \frac{15625}{64x^6}\).

Managing exponents correctly is essential not only to get the right powers of the variables but also to simplify each part of the term effectively so that we achieve the correct final expression.

Term Identification

Identifying the correct term in the binomial expansion follows from understanding both the structure of the expansion and the requirement of terms. In the binomial theorem, the \(k+1\)th term looks like:

• \(T_{k+1} = \binom{n}{k} a^{n-k} b^k\).

For example, if you're asked to find the 7th term, you should use \(k = 6\) since the \((k+1)\) notation helps us determine the exact term we're locating. Doing this allows us to plug the values directly into the formula. Each part of the process builds upon understanding how these parts work together to produce the entire expression or a single term within an expression.