Method of Differences (V-N Method): The Telescoping Series Shortcut

Master the Art of Collapsing Series for JEE Main

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1. Introduction

The Method of Differences, also called the V-N Method or Telescoping Summation, is a powerful technique for summing series where terms cancel in succession. It's frequently tested in JEE Main (1-2 questions/year) and transforms seemingly complex series into simple evaluations.

Core Insight: If each term can be expressed as $T_r = f(r) - f(r+1)$, then the sum "telescopes" to $f(1) - f(n+1)$.

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2. Fundamental Principle

2.1 Basic Telescoping

If $T_k = a_k - a_{k+1}$, then: $\sum_{k=1}^{n} T_k = (a_1 - a_2) + (a_2 - a_3) + \dots + (a_n - a_{n+1}) = a_1 - a_{n+1}$ Most intermediate terms cancel, leaving only the first and last.

2.2 Standard Form Recognition

The method applies when you can rewrite the general term as a difference. Common patterns include:

• Fractional terms with products in denominator
• Square root expressions (after rationalization)
• Factorial combinations
• Trigonometric terms using angle addition formulas

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3. Key Formulas and Techniques

3.1 Partial Fractions Approach

For $\frac{1}{P(r)Q(r)}$ where $P(r)$ and $Q(r)$ are linear and $Q(r) = P(r+1)$:

General Case:
If denominators form an arithmetic progression with common difference $d$: $\frac{1}{a_r a_{r+1}} = \frac{1}{d} \left( \frac{1}{a_r} - \frac{1}{a_{r+1}} \right)$ where $a_{r+1} - a_r = d$.

Examples:

1. $\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$ (here $d=1$)
2. $\frac{1}{(2r-1)(2r+1)} = \frac{1}{2} \left( \frac{1}{2r-1} - \frac{1}{2r+1} \right)$ (here $d=2$)

3.2 Higher Order Products

For three consecutive factors: $\frac{1}{a_r a_{r+1} a_{r+2}} = \frac{1}{2d} \left( \frac{1}{a_r a_{r+1}} - \frac{1}{a_{r+1} a_{r+2}} \right)$

3.3 Square Root Rationalization

$\frac{1}{\sqrt{a} + \sqrt{b}} = \frac{\sqrt{b} - \sqrt{a}}{b - a}$ If $b-a = d$ (constant), this becomes telescoping.

3.4 Factorial Series

$\frac{r}{(r+1)!} = \frac{1}{r!} - \frac{1}{(r+1)!}$ $\frac{r^2}{(r+1)!} = \frac{1}{(r-1)!} - \frac{1}{(r+1)!} \quad (r \ge 1)$

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4. Common Telescoping Patterns

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5. Step-by-Step Methodology

Step 1: Analyze the General Term

Identify if $T_r$ involves products, square roots, or factorials that suggest telescoping.

Step 2: Apply Appropriate Transformation

• For fractions: Use partial fractions
• For square roots: Rationalize
• For factorials: Use factorial identities
• For trigonometric: Use sum-to-product formulas

Step 3: Write Out First Few Terms

Verify the cancellation pattern by expanding: $S_n = [f(1)-f(2)] + [f(2)-f(3)] + \dots + [f(n)-f(n+1)]$

Step 4: Compute the Sum

$S_n = f(1) - f(n+1)$

Step 5: Simplify the Result

Express in simplest algebraic form.

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6. Worked Examples

Example 1: Basic Product Series

Problem: Sum of $\frac{1}{1\cdot 3} + \frac{1}{3\cdot 5} + \dots + \frac{1}{(2n-1)(2n+1)}$

Solution: $T_r = \frac{1}{(2r-1)(2r+1)} = \frac{1}{2} \left( \frac{1}{2r-1} - \frac{1}{2r+1} \right)$ $S_n = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) \right]$ $S_n = \frac{1}{2} \left( 1 - \frac{1}{2n+1} \right) = \frac{n}{2n+1}$

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Example 2: Square Root Series

Problem: $\frac{1}{1+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + \dots + \frac{1}{\sqrt{99}+\sqrt{100}}$

Solution: Rationalize each term: $\frac{1}{\sqrt{r}+\sqrt{r+1}} = \frac{\sqrt{r+1}-\sqrt{r}}{(r+1)-r} = \sqrt{r+1} - \sqrt{r}$ $S = (\sqrt{2}-\sqrt{1}) + (\sqrt{3}-\sqrt{2}) + \dots + (\sqrt{100}-\sqrt{99}) = \sqrt{100} - \sqrt{1} = 9$

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Example 3: Three-Factor Product

Problem: $\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \dots + \frac{1}{n(n+1)(n+2)}$

Solution: $\frac{1}{r(r+1)(r+2)} = \frac{1}{2} \left[ \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)} \right]$ $S_n = \frac{1}{2} \left[ \left( \frac{1}{1\cdot 2} - \frac{1}{2\cdot 3} \right) + \left( \frac{1}{2\cdot 3} - \frac{1}{3\cdot 4} \right) + \dots \right]$ $S_n = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{(n+1)(n+2)} \right) = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}$

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7. JEE Main PYQs with Detailed Solutions

PYQ 1 (JEE Main 2021)

Problem: $\frac{1}{3^2-1} + \frac{1}{5^2-1} + \dots + \frac{1}{201^2-1} = ?$

Solution: $T_r = \frac{1}{(2r+1)^2-1} = \frac{1}{4r(r+1)} = \frac{1}{4} \left( \frac{1}{r} - \frac{1}{r+1} \right)$ Here $2r+1$ goes from 3 to 201 ⇒ $r$ from 1 to 100. $S = \frac{1}{4} \left[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{100} - \frac{1}{101}\right) \right]$ $S = \frac{1}{4} \left( 1 - \frac{1}{101} \right) = \frac{25}{101}$

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PYQ 2 (JEE Main 2023)

Problem: Given $a_5 = 2a_7$ and $a_{11} = 18$ for an AP. Find $12\sum_{k=10}^{17} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$

Solution: Let AP be $a_n = a_1 + (n-1)d$.

From $a_5 = 2a_7$: $a_1 + 4d = 2(a_1 + 6d) ⇒ a_1 = -8d$
From $a_{11} = 18$: $-8d + 10d = 18 ⇒ d = 9, a_1 = -72$

Rationalize: $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$ $\sum_{k=10}^{17} = \frac{1}{d} (\sqrt{a_{18}} - \sqrt{a_{10}}) = \frac{1}{9} (\sqrt{81} - \sqrt{9}) = \frac{6}{9} = \frac{2}{3}$ Multiply by 12: $12 \times \frac{2}{3} = 8$

Answer: 8

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PYQ 3 (JEE Main 2019)

Problem: Sum of first 10 terms of $\frac{3}{1^2\cdot 2^2} + \frac{5}{2^2\cdot 3^2} + \frac{7}{3^2\cdot 4^2} + \dots$

Solution: $T_r = \frac{2r+1}{r^2(r+1)^2} = \frac{1}{r^2} - \frac{1}{(r+1)^2}$ $S_{10} = \left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \dots + \left(\frac{1}{100} - \frac{1}{121}\right) = 1 - \frac{1}{121} = \frac{120}{121}$

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8. Advanced Applications

8.1 Trigonometric Telescoping

Using product-to-sum identities: $\sum_{k=1}^{n} \sin(kx) \sin((k+1)x) = \frac{1}{2} \sum_{k=1}^{n} [\cos x - \cos((2k+1)x)]$ This can telescope if further manipulated.

8.2 Combined AP-GP Series

Sometimes rewriting helps: $\sum_{r=1}^{n} r \cdot a^{r-1} = \frac{d}{da} \left( \sum_{r=1}^{n} a^{r-1} \right)$ Differentiate the geometric series formula.

8.3 Nested Summations

For double sums, interchange order or use: $\sum_{i=1}^{n} \sum_{j=1}^{i} f(j) = \sum_{j=1}^{n} (n-j+1) f(j)$

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9. Special Cases and Tricks

9.1 When Denominators Aren't Consecutive

For $\frac{1}{r(r+k)}$ where $k>1$: $\frac{1}{r(r+k)} = \frac{1}{k} \left( \frac{1}{r} - \frac{1}{r+k} \right)$ The sum doesn't fully telescope but has only $k$ surviving terms.

9.2 Sum of Reciprocals of Products of AP Terms

General formula for $a_r = a + (r-1)d$: $\sum_{r=1}^{n} \frac{1}{a_r a_{r+1} \dots a_{r+m-1}} = \frac{1}{(m-1)d} \left( \frac{1}{a_1 a_2 \dots a_{m-1}} - \frac{1}{a_{n+1} a_{n+2} \dots a_{n+m-1}} \right)$

9.3 Using Known Sums

Sometimes it's easier to compute $\sum_{r=1}^{n} T_r$ by finding $\sum_{r=1}^{n} r$, $\sum_{r=1}^{n} r^2$, etc., and combining.

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10. Common Errors and Verification Tips

Common Mistakes:

1. Incorrect partial fractions: Always verify by recombining.
2. Misidentifying $d$: Ensure you compute $a_{r+1} - a_r$ correctly.
3. Off-by-one errors: Check first and last terms explicitly.
4. Assuming telescoping when it doesn't occur: Write out 3-4 terms to verify cancellation.

Verification:

• For small $n$ (like $n=3$), compute both direct sum and telescoped form.
• Check limiting behavior as $n \to \infty$ (if infinite series).
• Ensure final expression has correct dimensions/units.

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11. Practice Problems

Level 1:

1. $\sum_{r=1}^{n} \frac{1}{r(r+3)}$
2. $\sum_{r=1}^{n} \frac{1}{\sqrt{r} + \sqrt{r+2}}$

Level 2: 3. $\sum_{r=1}^{n} \frac{r}{(r+1)!}$ 4. $\sum_{r=1}^{n} \frac{2r+1}{r^2(r+1)^2}$

Level 3 (JEE): 5. If $S_n = \sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)}$, find $\lim_{n\to\infty} S_n$ 6. $\sum_{r=1}^{n} \frac{3r^2 - 1}{(r^2 - r + 1)(r^2 + r + 1)}$

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12. Quick Reference Guide

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13. Final Strategy for JEE Main

1. Recognize the pattern within 30 seconds.
2. Apply partial fractions or rationalization immediately.
3. Write 2-3 terms to confirm telescoping.
4. Compute sum using first minus last surviving term.
5. Simplify and box the answer.

Time Allocation: Such problems should take 2-3 minutes maximum.

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Mastering telescoping series turns one of the most daunting topics into a quick, reliable point-scorer. Practice until recognition becomes instantaneous!