Problem 81 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\sum_{r=1}^{n} r x^{r-1}=\frac{1}{(1-x)^{2}} \cdot\left\\{1+a x^{n}+b x^{n+1}\right\\}\), then (A) \(a=(n+1)\) (B) \(b=n\) (C) \(a=-(n+1)\) (D) \(b=-n\)

Step-by-Step Solution

Answer

The correct option is (C) \(a=-(n+1)\).

Step 1: Understand the Problem

The problem provides a series sum in terms of powers of \(x\) and equates it to an expression involving fractions and polynomial terms with coefficients \(a\) and \(b\). We need to determine the correct values of \(a\) and \(b\).

Step 2: Express the Given Series

The problem states \( \sum_{r=1}^{n} r x^{r-1} = \frac{1}{(1-x)^{2}} \cdot \left\{1 + ax^n + b x^{n+1}\right\} \). Consider the left side as a derivative of basic geometric series \( \sum_{r=0}^{n} x^r = \frac{1-x^{n+1}}{1-x} \).

Step 3: Derive the Expression for Series

The series \( \sum_{r=1}^{n} r x^{r-1} \) is the derivative of \( \sum_{r=1}^{n} x^r \) with respect to \(x\), evaluated as \( \frac{d}{dx}[\sum_{r=1}^{n} x^r] = \sum_{r=1}^{n} r x^{r-1} \). This derivative is equal to \( \frac{d}{dx}(\frac{1-x^{n+1}}{1-x}) \).

Step 4: Differentiate the Geometric Series

Take the derivative: \( \frac{d}{dx} \left[ \frac{1-x^{n+1}}{1-x} \right] = \frac{(n+1)x^n}{1-x} + \frac{1-x^{n+1}}{(1-x)^2} \). This matches the left side of the original equation.

Step 5: Combine and Simplify Expressions

The derivative simplifies to \( \frac{1}{(1-x)^2} - \frac{(n+1)x^n}{1-x} \). Both expressions have a common denominator \( (1-x)^2 \), giving \( \frac{1}{(1-x)^2} = \frac{1 - (n+1)x^n}{(1-x)^2} + ax^n + bx^{n+1} \).

Step 6: Match Coefficients of Equal Terms

Compare terms from both sides to determine \(a\) and \(b\): \( 1-x^n - (n+1)x^n \equiv 1 + a x^n + b x^{n+1}\). This shows that \(a = -(n+1)\) and since there is no \(x^{n+1}\) in \(1-(n+1)x^n\), \(b = 0\).

Step 7: Check Possible Answers

From the coefficients found, \(a=-(n+1)\) matches option (C), and \(b=0\), which is not provided in the options, but the other option \(b=-n\) doesn't fit.

Key Concepts

Series Differentiation

Series differentiation is a powerful tool in mathematical analysis. It allows us to find the derivative of a series, often leading to a more insightful understanding of its behavior.
When dealing with series differentiation, we look at the summation of expressions and perform differentiation term by term.

• Differentiating a series involves transforming each term in the series based on its position and mathematical form.
• This can simplify the problem-solving process, especially when faced with complex algebraic expressions.

In the context of the given exercise, the series \[ \sum_{r=1}^{n} r x^{r-1} \]is the derivative of a geometric series, showing how we can apply calculus to understand patterns and relationships within series.
By applying differentiation to the geometric series,\[ \sum_{r=0}^{n} x^r = \frac{1-x^{n+1}}{1-x} \], we simplify the task of finding the derivative and further explore its algebraic structure.

Geometric Progression

A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
In mathematical terms, the progression can be expressed as:\[ a, ar, ar^2, ar^3, \, \ldots \]

• Here, \(a\) represents the first term, and \(r\) is the common ratio.
• This kind of progression is characterized by a constant ratio between successive terms.

In this exercise, the series \[ \sum_{r=1}^{n} r x^{r-1} \] is linked to a geometric series form. Understanding how to manipulate and differentiate geometric series helps solve and simplify expressions in calculations and scientific analyses. This geometric progression plays a pivotal role in deriving complex mathematical series and finding their sums.

Expression Simplification

Expression simplification is the process of reducing complex mathematical expressions into simpler, more manageable forms. This process involves algebraic manipulation, where terms are combined, like terms are merged, and unnecessary components are eliminated.

• It makes expressions easier to work with and can often reveal fundamental properties of the equation.
• It often involves finding common denominators, factorizations, and combining like terms.

For the given problem, after differentiating, the expression simplifies to\[ \frac{1}{(1-x)^2} - \frac{(n+1)x^n}{1-x} \].
By finding a common denominator,this expression becomes easier to work with, especially when matching coefficients to find specific values for \(a\) and \(b\). This process demonstrates how simplification allows us to compare equations and solve for unknowns efficiently. Whether in algebra, calculus, or real-world problem-solving, simplifying complex expressions underpins our ability to extract meaningful and actionable information.