Binomial Theorem for Any Index | JEE Advanced Concept

Beyond Positive Integers: A Revolutionary Extension

Until now, we've worked with the binomial theorem for positive integer indices. But what if we want to expand $(1 + x)^{1/2}$ or $(1 + x)^{-3}$? This is where the general binomial theorem comes in - one of the most powerful extensions in algebra.

🎯 Why This Matters for JEE Advanced

Appears in approximation problems and limit calculations
Essential for understanding infinite series and power series
Used in calculus for Taylor series expansions
Tested in comprehension-type questions in JEE Advanced

1. The General Binomial Theorem

Binomial Theorem for Any Rational Index

$$ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots $$

Where:

$n$ is any rational number (n ∈ Q)
The series is infinite
General term: $T_{r+1} = \frac{n(n-1)\cdots(n-r+1)}{r!}x^r$
Valid when |x| < 1

Special Cases:

If $n$ is positive integer: finite series
If $n$ is negative integer: infinite series
If $n$ is fraction: infinite series
If $n = -1$: geometric series

💡 Understanding the General Term

The general term in the expansion of $(1 + x)^n$ for any rational $n$ is:

$$ T_{r+1} = \binom{n}{r} x^r $$

Where the binomial coefficient is defined as:

$$ \binom{n}{r} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!} $$

Note: This definition works for any real $n$, not just positive integers!

2. The Crucial Validity Condition: |x| < 1

⚠️ Important Restriction

The infinite binomial expansion converges (gives a finite sum) only when:

$$ |x| < 1 $$

If $|x| \geq 1$, the series diverges (the sum becomes infinite or undefined).

🔍 Why |x| < 1 is Necessary

Mathematical Reason:

The terms of the series must eventually decrease in magnitude for the sum to converge to a finite value.

When $|x| < 1$, higher powers of $x$ become smaller, making the series convergent.

Visual Understanding:

Consider $(1 + x)^{-1} = 1 - x + x^2 - x^3 + x^4 - \cdots$

If $x = 0.5$: Terms: 1, -0.5, 0.25, -0.125, 0.0625, ... (converges to 2/3)

If $x = 2$: Terms: 1, -2, 4, -8, 16, ... (diverges to infinity)

🎯 JEE Advanced Insight

For $(a + x)^n$ where $a \neq 1$, rewrite as:

$$ (a + x)^n = a^n \left(1 + \frac{x}{a}\right)^n $$

Then the validity condition becomes $\left|\frac{x}{a}\right| < 1$ or $|x| < |a|$.

3. The Concept of Infinite Series

∞ Understanding Infinite Series

An infinite series is a sum of infinitely many terms:

$$ S = a_1 + a_2 + a_3 + a_4 + \cdots $$

Convergent Series

The sum approaches a finite limit.

Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 2$

Divergent Series

The sum grows without bound or oscillates.

Example: $1 + 2 + 3 + 4 + \cdots = \infty$

📚 Example: Expansion of (1 + x)^{1/2}

Let's expand $\sqrt{1 + x} = (1 + x)^{1/2}$ using the general binomial theorem:

Step 1: Apply the formula with n = 1/2

$(1 + x)^{1/2} = 1 + \frac{1}{2}x + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}x^2 + \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}x^3 + \cdots$

Step 2: Simplify coefficients

$= 1 + \frac{1}{2}x + \frac{-\frac{1}{4}}{2}x^2 + \frac{\frac{3}{8}}{6}x^3 + \cdots$

$= 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots$

Step 3: Validity condition

This expansion is valid when $|x| < 1$.

4. Important Standard Expansions

(1 + x)^{-1}

$$ (1 + x)^{-1} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots $$

Valid for $|x| < 1$

This is the geometric series with ratio $-x$

(1 + x)^{-2}

$$ (1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + 5x^4 - \cdots $$

Valid for $|x| < 1$

(1 - x)^{-1}

$$ (1 - x)^{-1} = 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots $$

Valid for $|x| < 1$

(1 + x)^{1/2}

$$ (1 + x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots $$

Valid for $|x| < 1$

5. Applications & Problem Solving

🎯 JEE Advanced Type Problem

Problem: Find the coefficient of $x^3$ in the expansion of $\frac{1}{\sqrt{1 - 2x}}$ up to $x^3$.

Step 1: Rewrite the expression

$\frac{1}{\sqrt{1 - 2x}} = (1 - 2x)^{-1/2}$

Step 2: Apply binomial theorem with n = -1/2

$(1 - 2x)^{-1/2} = 1 + (-\frac{1}{2})(-2x) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-2x)^2 + \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}(-2x)^3 + \cdots$

Step 3: Simplify each term

$= 1 + x + \frac{3}{8}(4x^2) + \frac{-15}{48}(-8x^3) + \cdots$

$= 1 + x + \frac{3}{2}x^2 + \frac{5}{2}x^3 + \cdots$

Step 4: Check validity condition

$| -2x | < 1 \Rightarrow |x| < \frac{1}{2}$

Final Answer

Coefficient of $x^3$ is $\frac{5}{2}$

💡 Approximation Technique

The binomial expansion is extremely useful for approximations when $x$ is small:

$$ (1 + x)^n \approx 1 + nx \quad \text{(for small } x \text{)} $$

Example: $\sqrt{1.02} = (1 + 0.02)^{1/2} \approx 1 + \frac{1}{2}(0.02) = 1.01$

The actual value is approximately 1.00995, so our approximation is quite good!

6. Common Mistakes to Avoid

❌ Forgetting the Validity Condition

Using the expansion when $|x| \geq 1$ leads to incorrect results.

Wrong: Applying $(1 + 2)^{-1} = 1 - 2 + 4 - 8 + \cdots$ (diverges!)

Correct: $(1 + 2)^{-1} = \frac{1}{3}$ (evaluate directly)

❌ Misapplying to (a + b)^n

For $(a + b)^n$ where $a \neq 1$, you must factor:

$$ (a + b)^n = a^n\left(1 + \frac{b}{a}\right)^n $$

Then the validity condition is $\left|\frac{b}{a}\right| < 1$.

❌ Confusing Finite and Infinite Cases

Remember: The expansion is finite only when $n$ is a positive integer.

For all other rational $n$, the expansion is infinite.

📋 Quick Revision Checklist

Key Formulas:

$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots$
$T_{r+1} = \frac{n(n-1)\cdots(n-r+1)}{r!}x^r$
Valid for $|x| < 1$ when $n \notin \mathbb{Z}^+$
For $(a + x)^n = a^n(1 + \frac{x}{a})^n$

Must Remember:

$(1 + x)^{-1} = 1 - x + x^2 - x^3 + \cdots$
$(1 - x)^{-1} = 1 + x + x^2 + x^3 + \cdots$
$(1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \cdots$
$(1 + x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \cdots$

📝 Practice Problems

Test your understanding with these JEE-level problems:

1. Find the first four terms in the expansion of $(1 - 3x)^{-1/2}$

State the condition for validity.

2. Find the coefficient of $x^4$ in $(1 + 2x)^{-3}$

Use the general binomial theorem.

3. Using binomial theorem, approximate $\sqrt[3]{1.03}$ to 4 decimal places

Hint: Write as $(1 + 0.03)^{1/3}$

Ready for More Advanced Concepts?

Continue your journey through the Binomial Theorem series

Previous: Multinomial Theorem Next: Applications & Problem Solving

Beyond Positive Integers: A Revolutionary Extension

🎯 Why This Matters for JEE Advanced

🧭 Navigation Guide

1. The General Binomial Theorem

Binomial Theorem for Any Rational Index

Where:

Special Cases:

💡 Understanding the General Term

2. The Crucial Validity Condition: |x| < 1

⚠️ Important Restriction

🔍 Why |x| < 1 is Necessary

Mathematical Reason:

Visual Understanding:

🎯 JEE Advanced Insight

3. The Concept of Infinite Series

∞ Understanding Infinite Series

Convergent Series

Divergent Series

📚 Example: Expansion of (1 + x)^{1/2}

Step 1: Apply the formula with n = 1/2

Step 2: Simplify coefficients

Step 3: Validity condition

4. Important Standard Expansions

(1 + x)^{-1}

(1 + x)^{-2}

(1 - x)^{-1}

(1 + x)^{1/2}

5. Applications & Problem Solving

🎯 JEE Advanced Type Problem

Step 1: Rewrite the expression

Step 2: Apply binomial theorem with n = -1/2

Step 3: Simplify each term

Step 4: Check validity condition

Final Answer

💡 Approximation Technique

6. Common Mistakes to Avoid

❌ Forgetting the Validity Condition

❌ Misapplying to (a + b)^n

❌ Confusing Finite and Infinite Cases

📋 Quick Revision Checklist

Key Formulas:

Must Remember:

📝 Practice Problems

Ready for More Advanced Concepts?