Problem 3 - Algebra for JEE ADVANCED

Question

The value of \(\sum_{r=0}^{n-1}{\underline{\phantom{xx}}}^{n} C_{r} /\left({ }^{n} C_{r}+{ }^{n} C_{r+1}\right)\) equals a. \(n+1\) b. \(n / 2\) c. \(n+2\) d. none of these

Step-by-Step Solution

Answer

The answer is b. \( \frac{n}{2} \).

Step 1: Understand the Expression

We are given the expression \( \sum_{r=0}^{n-1} \frac{{C(n, r)}}{C(n, r) + C(n, r+1)} \). Here \(C(n, r)\) represents the binomial coefficient, calculated as \( \frac{n!}{r!(n-r)!} \). We need to find the sum of this expression from \( r=0 \) to \( n-1 \).

Step 2: Simplify Individual Terms

Let's focus on simplifying a single term: \( \frac{{C(n, r)}}{C(n, r) + C(n, r+1)} \). Using the property \( C(n, r+1) = \frac{n-r}{r+1} C(n, r) \), rewrite the denominator as \( C(n, r) + \frac{n-r}{r+1} C(n, r) = C(n, r) \left(1 + \frac{n-r}{r+1}\right) \).

Step 3: Factor and Simplify Term

The denominator can be factored as \( C(n, r) \cdot \frac{n+1}{r+1} \). Thus, the expression for each term becomes \( \frac{{C(n, r)}}{C(n, r) \cdot \frac{n+1}{r+1}} = \frac{r+1}{n+1} \).

Step 4: Evaluate the Sum

Now, we replace the terms in the summation with our simplified form: \( \sum_{r=0}^{n-1} \frac{r+1}{n+1} \). This sum simplifies to \( \frac{1}{n+1} \sum_{r=0}^{n-1} (r+1) \).

Step 5: Sum of Arithmetic Series

The inner summation, \( \sum_{r=0}^{n-1} (r+1) \), is an arithmetic series which equals \( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \). Substitute this back in to get \( \frac{1}{n+1} \cdot \frac{n(n+1)}{2} \).

Step 6: Final Simplification

Cancel out \( n+1 \) in the numerator and denominator to get \( \frac{n}{2} \). Therefore the sum simplifies to \( \frac{n}{2} \).

Step 7: Verify and Conclude

The calculated sum is \( \frac{n}{2} \), which matches option b. This confirms we have followed the steps correctly and the sum simplifies correctly.

Key Concepts

Summation

Summation involves the addition of a sequence of numbers or expressions. In mathematics, summation is represented by the sigma notation \( \Sigma \), which allows us to compactly express the sum of a series of terms. For instance, \( \sum_{r=0}^{n-1} \frac{C(n, r)}{C(n, r) + C(n, r+1)} \) instructs us to add up the terms derived from this formula as \(r\) takes each integer value from 0 to \(n-1\). Each term in our sum consists of complex expressions involving binomial coefficients, making clear steps and simplifications crucial to obtaining the final result.

• Start by identifying the structure of the summation, noting how many terms need to be added.
• Understand the individual components of each term before addressing the entire series.

Grasping the full picture of what is being summed is often the first challenge in these exercises. Breaking down each element is vital before simplifying and evaluating the sum.

Arithmetic Series

An arithmetic series is the sum of terms in an arithmetic sequence, where each term after the first is generated by adding a fixed number, called the common difference. In the exercise, we came across the series \( \sum_{r=0}^{n-1} (r+1) \). Since \(r+1\) represents a sequential counting from 1 to \(n\), this is a classic example of an arithmetic series.

• The sum of the first \(n\) positive integers is given by the formula \( \frac{n(n+1)}{2} \).
• This is derived by recognizing the pattern in the consecutive number addition and expressing it as a single formula.

Arithmetic series are straightforward to work with once you identify the pattern or sequence, and their total can be quickly computed with the formula provided, allowing for simplified calculations in larger algebraic expressions.

Simplification of Expressions

Simplifying expressions is about reducing them into an easier or more concise form without changing their value. During this exercise, simplification began by analyzing each term of the series: \( \frac{C(n, r)}{C(n, r) + C(n, r+1)} \).

• First, use properties of binomial coefficients to transform and reduce the terms, such as \( C(n, r+1) = \frac{n-r}{r+1} C(n, r) \).
• Factor common elements in the denominator to simplify the fraction.
• After simplifying the expression for individual terms, address the entire sum, ensuring all operations and simplifications maintain mathematical integrity.

By simplifying expressions, complex problems become manageable, allowing you to solve them step-by-step with clarity and confidence. It's about recognizing patterns, utilizing algebraic properties, and ultimately achieving clear solutions to what may initially appear convoluted.