Problem 88 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the expansion in powers of \(x\) of the function \(\frac{1}{(1-a x)(1-b x)}\) is \(a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3} \ldots .\) then \(a_{n}\) is [2006] (a) \(\frac{b^{n}-a^{n}}{b-a}\) (b) \(\frac{a^{n}-b^{n}}{b-a}\) (c) \(\frac{a^{n+1}-b^{n+1}}{b-a}\) (d) \(\frac{b^{n+1}-a^{n+1}}{b-a}\)

Step-by-Step Solution

Answer

The correct answer is (c) \(\frac{a^{n+1}-b^{n+1}}{b-a}\).

Step 1: Express the function as a product of two geometric series

The function \( \frac{1}{(1-ax)(1-bx)} \) can be expressed as the product of two geometric series: \( \frac{1}{1-ax} \) and \( \frac{1}{1-bx} \). Each of these series can be individually expanded as: \[ \frac{1}{1-ax} = 1 + ax + a^2x^2 + a^3x^3 + \ldots \]\[ \frac{1}{1-bx} = 1 + bx + b^2x^2 + b^3x^3 + \ldots \]

Step 2: Multiply the series

Multiply the two series together to obtain the series expansion of the function. This involves expanding:\((1 + ax + a^2x^2 + a^3x^3 + \ldots)(1 + bx + b^2x^2 + b^3x^3 + \ldots)\).Consider only the terms up to the desired power of \( x \), which contribute to \( a_n \).

Step 3: Compute the coefficient of x^n

The coefficient of \( x^n \) in the expanded series is given by the sum of all products of coefficients from both series such that the powers of \( a \) and \( b \) add up to \( n \). That is, \[ \sum_{k=0}^{n} a^k b^{n-k} \].

Step 4: Use the binomial theorem

The sum \( \sum_{k=0}^{n} a^k b^{n-k} \) is a finite geometric series. It can be simplified using the formula for the sum of a geometric series:\[ \sum_{k=0}^{n} c^k = \frac{c^{n+1}-1}{c-1} \]For our case, the formula becomes:\[ \sum_{k=0}^{n} a^k b^{n-k} = \frac{a^{n+1} - b^{n+1}}{a - b} \] if \( a eq b \).

Step 5: Choose the correct answer

Comparing the expression for the coefficient \( a_n \) with the given options, we find that it matches option (c): \[ \frac{a^{n+1}-b^{n+1}}{b-a} \].

Key Concepts

Geometric Series

A geometric series is a series with a constant ratio between successive terms. It's foundational for many mathematical concepts and is present in everyday calculations.
Consider the series in which each term gets multiplied by a fixed number to yield the next term, for example, in the sequence 1, 2, 4, 8, the factor is 2. The general form of a geometric series is:

• The first term: \( a \)
• Common ratio: \( r \)
• Sum of the first \( n \) terms: \( S_n = a \frac{1-r^n}{1-r} \)

In the exercise, the expressions \( \frac{1}{1-ax} \) and \( \frac{1}{1-bx} \) are expanded using geometric series. This means representing them as indefinite sums to simplify multiplication, allowing for a comprehensive series expansion later.

Binomial Theorem

The binomial theorem provides an excellent way to expand expressions that are raised to any given power. Its formula aids in expanding the power of sums and involves combinations.For any positive integer \( n \), the binomial theorem states:

• \((x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \)

This allows the expression of powers of sums without the need to individually calculate each element. In the context of our problem, the binomial theorem wasn't directly used for expansion, but its principle helps simplify the process by providing a framework for understanding geometric series expansion in polynomials.

Series Expansion

Series expansion is a powerful mathematical tool allowing functions to be expressed as a sum of terms calculated from the function's derivatives at a single point. This technique is very useful for simplifying complex functions and operations.Series expansion makes use of both Taylor and Maclaurin series for functions. Our exercise demonstrates the one common use-case of series expansion for rational functions using geometric series, expressed as:

• \( \frac{1}{1-ax} \) as \( 1 + ax + a^2x^2 + a^3x^3 + \ldots \)
• \( \frac{1}{1-bx} \) similarly expanded.

This allows us to transform the original complex expression \( \frac{1}{(1-ax)(1-bx)} \) into manageable parts. By multiplying these series, we form a combined series that reflects the coefficients of \( x^n \), yielding the solution.