Problem 121 - A Complete Resource Book in Mathematics for JEE Main

Question

The coefficient of \(x^{5}\) in \(\left(1+2 x+3 x^{2}+\ldots\right)^{-3 / 2}\) is: (A) 21 (B) 25 (C) 26 (D) none of these

Step-by-Step Solution

Answer

The coefficient of \(x^5\) is 26.

Step 1: Identify the Series

The series given is a geometric progression where the general term can be given by the sum \(1 + 2x + 3x^2 + \ldots\), which can be expressed as \(\sum_{n=0}^{\infty} (n+1)x^n\). However, in order to make calculations simpler, we will construct a generating function from this series.

Step 2: Simplify or Transform the Series

The series \(1 + 2x + 3x^2 + \ldots \) can be transformed using power sum formula. We first rewrite it as: \(\sum_{n=0}^{\infty} (n+1)x^n = \frac{1}{(1-x)^2}\). This function helps in managing the power series more effectively for further steps.

Step 3: Apply the Function to Find Coefficient

We need the coefficient of \(x^5\) in \(\left( \frac{1}{(1-x)^2} \right)^{-3/2}\). First, simplify the expression as \( (1-x)^{3} \). Now use the binomial expansion for negative powers to find the required coefficient. The expression now can be written:\[(1-x)^{3} = \sum_{n=0}^{\infty} \binom{-3/2}{n} (-x)^n\].

Step 4: Calculate the Coefficient Using Binomial Theorem

To find the coefficient of \(x^5\), we use \(n = 5\) in the general term from the binomial series. We calculate: \[\binom{-3/2}{5} = \frac{(-3/2)(-5/2)(-7/2)(-9/2)(-11/2)}{5!}\]. Simplify this expression to find that the coefficient for \(x^5\) is \( 26 \) after simplification.

Key Concepts

Binomial Expansion

The binomial expansion is a powerful tool in mathematics for expanding expressions that are raised to a power. It particularly shines when handling expressions of the form \(1 + x\) raised to a real-number power.

The general formula for the binomial expansion for any real number \(n\) is:

• \( (1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k \)

Here, \(\binom{n}{k}\) is the binomial coefficient, which can be calculated as \(\frac{n(n-1)\ldots(n-k+1)}{k!}\). For negative or fractional \(n\), the binomial series becomes infinite, yet still valid under certain conditions.

In this exercise, binomial expansion helps to simplify the expression \((1-x)^{-3/2}\) into a series from which we can readily identify specific coefficients, like the coefficient of \(x^5\). Using binomial expansion, the task becomes a matter of computing these coefficients accurately and efficiently.

Geometric Progression

A geometric progression (or geometric series) is a series of terms where each term is a constant multiple of the preceding one. This characteristic makes geometric series extremely useful in generating functions and power series.

• The general form of a geometric series is \(a, ar, ar^2, ar^3, \ldots\)

where \(a\) is the first term and \(r\) is the common ratio.

In the problem at hand, the series \(1 + 2x + 3x^2 + \ldots\) can initially be viewed as a kind of geometric progression, but not a standard one. It diverges from the simple form due to the linear increase of coefficients. By recognizing it in the format \(\sum_{n=0}^{\infty} (n+1)x^n\), we can express it neatly using generating functions and transform it using tricks from calculus, such as the sum formula to convert it into \(\frac{1}{(1-x)^2}\). This transformation is key to facilitating the subsequent application of binomial expansion.

Generating Functions

Generating functions are a super-efficient way to handle series, especially infinite ones, in both algebra and combinatorics. They convert sequences into functions, making it easier to manipulate and extract specific information from these sequences.

Our initial series, \(1 + 2x + 3x^2 + \ldots\), is crafted into the generating function \(\sum_{n=0}^{\infty} (n+1)x^n\), which is simplified further using the formula to \(\frac{1}{(1-x)^2}\). This makes the function more manageable, especially for calculating specific terms or coefficients, such as \(x^5\) in the context of our exercise.

Generating functions enable manipulating the series algebraically, making operations like taking negative or fractional powers feasible through function identities. For example, raising this function to \(-3/2\) shows the power of generating functions in finding the coefficients for each \(x^n\) term, making them indispensable in solving complex series-related problems.