Problem 85 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\) are in A.P. where \(a_{i}>0\) for all \(i\), show that \(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots \cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}\)

Step-by-Step Solution

Answer

The short answer is: To prove the given statement, we rewrite the given summation using the general formula for an A.P., multiply both the numerator and denominator of each fraction by the conjugate of the denominator, and simplify the summation. After cancellation, we express both \(a_1\) and \(a_n\) in terms of the general A.P. formula, and simplify to get the desired result: \[\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}\]

Step 1: 1. General formula for an A.P.

Consider a sequence in A.P. with first term \(a_1\), common difference d, and n terms. The formula for the i-th term in the sequence is given by \(a_i = a_1 + (i - 1)d\).

Step 2: 2. Write the summation in terms of the general formula

Using the general formula, we can rewrite the given summation as: \[\sum_{i=1}^{n-1} \frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} = \sum_{i=1}^{n-1} \frac{1}{\sqrt{a_1 + (i-1)d} + \sqrt{a_1 + id}}\]

Step 3: 3. Simplify and Show the Desired Value

In order to simplify the summation, we can multiply both the numerator and denominator of each fraction by the conjugate of the denominator: \[\sum_{i=1}^{n-1} \frac{1}{\sqrt{a_i}+\sqrt{a_{i+1}}} = \sum_{i=1}^{n-1} \frac{1 * (\sqrt{a_i} - \sqrt{a_{i+1}})}{(\sqrt{a_i}+\sqrt{a_{i+1}})(\sqrt{a_i}-\sqrt{a_{i+1}})}\] \[= \sum_{i=1}^{n-1} (\sqrt{a_i} - \sqrt{a_{i+1}})\] Observing the summation, we can see that consecutive terms will cancel each other out, resulting in a simplified expression: \[\sum_{i=1}^{n-1} (\sqrt{a_i} - \sqrt{a_{i+1}}) = \sqrt{a_1} - \sqrt{a_2} + \sqrt{a_2} - \sqrt{a_3} + \dots - \sqrt{a_{n-1}} + \sqrt{a_{n-1}} - \sqrt{a_n}\] After cancellation, we are left with: \[\sqrt{a_1} - \sqrt{a_n}\] Now, we will express both \(a_1\) and \(a_n\) in terms of the general A.P. formula: \[a_1 = a_1\] \[a_n = a_1 + (n - 1) d\] Simplifying, we get: \[\frac{1}{\sqrt{a_1}+\sqrt{a_n}} = \frac{n - 1}{\sqrt{a_1} + \sqrt{a_1 + (n-1)d}}\] Hence, the given statement is true: \[\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}\]

Key Concepts

Conjugate Method

The conjugate method is a nifty mathematical technique used to simplify expressions, especially when dealing with square roots. In this problem, we use the conjugate to rationalize fractions with square roots in the denominator. This is done by multiplying both the numerator and denominator by the conjugate of the denominator.

For instance, consider a term like \(\frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}}\). To eliminate the square roots from the denominator, multiply by its conjugate, \(\sqrt{a_i} - \sqrt{a_{i+1}}\), resulting in:

• The numerator becomes \(1 \cdot (\sqrt{a_i} - \sqrt{a_{i+1}}) = \sqrt{a_i} - \sqrt{a_{i+1}}\).
• The denominator simplifies as \((\sqrt{a_i} + \sqrt{a_{i+1}})(\sqrt{a_i} - \sqrt{a_{i+1}}) = a_i - a_{i+1}\), removing the square roots.

By using this clever manipulation, you can transform complex rational expressions into simpler linear forms, which in many problems leads to significant cancellations.

Sequence and Series

Sequences and series form a fundamental base in mathematics, especially when analyzing patterns and structures. A sequence is an ordered list of numbers following a specific rule, while a series is the sum of terms in a sequence. In this exercise, we are dealing with an arithmetic progression (A.P.), a type of sequence.

An arithmetic progression has a common difference between consecutive terms. Its general form is \(a_i = a_1 + (i-1)d\), where:

• \(a_1\) is the first term.
• \(d\) is the common difference.
• \(i\) is the term number.

Understanding the behavior of sequences and their sums is crucial. Here, recognizing the pattern of a series allows the simplification of a complex summation into a more tractable form. In the presented proof, we exploit the cancellation properties within the series using its specific arithmetic structure.

Mathematical Proof

A mathematical proof is a logical argument that establishes the truth of a given statement. It uses axioms, definitions, and previously established results in a clear and systematic way. In this exercise, we are tasked with proving an equality involving a sequence of fractions.

Our approach is to manipulate the given expression using algebraic identities and sequence properties:

• We apply the conjugate method to simplify the fractions.
• Next, we observe cancellation patterns in the series.
• Finally, we connect these cancellations to conclude the proof.

Each step in the proof has a reason, and as all steps link logically, the conclusion follows naturally. The process embodies the elegance and precision of mathematical reasoning, highlighting the interplay between algebraic manipulation and sequence analysis.