Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)**
Published: October 2025
Category: Algebra | Binomial Expansion | Advanced Topics

🔹 Introduction

The Binomial Theorem is one of the most powerful results in algebra — it expands expressions of the form (1+x)n(1 + x)^n(1+x)n for any exponent nnn.

Most students learn it for positive integers like n=3n = 3n=3 or 555, but what happens when nnn is a fraction or a negative number?

That’s where the general form of the binomial theorem comes in — a key concept for advanced mathematics and series expansions.

⚡️ 1️⃣ The Standard Binomial Theorem

For any positive integer nnn: (1+x)n=1+(n1)x+(n2)x2+⋯+(nn)xn(1 + x)^n = 1 + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n(1+x)n=1+(1n​)x+(2n​)x2+⋯+(nn​)xn

Where the binomial coefficient is: (nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n – r)!}(rn​)=r!(n−r)!n!​

This expansion has finite terms because nnn is an integer.

Example: (1+x)3=1+3x+3×2+x3(1 + x)^3 = 1 + 3x + 3x^2 + x^3(1+x)3=1+3x+3×2+x3

🌟 2️⃣ Extending to Fractional or Negative Powers

When nnn is not a positive integer — say a fraction or negative integer — the series becomes infinite.

The formula still holds if we redefine (nr)\binom{n}{r}(rn​) for all real (or even complex) values of nnn: (nr)=n(n−1)(n−2)⋯(n−r+1)r!\boxed{\displaystyle \binom{n}{r} = \frac{n(n – 1)(n – 2)\cdots(n – r + 1)}{r!}}(rn​)=r!n(n−1)(n−2)⋯(n−r+1)​​

This definition works for any real or fractional nnn, even if n<0n < 0n<0.

🧩 3️⃣ Generalized Binomial Theorem

For any real number nnn and ∣x∣<1|x| < 1∣x∣<1: (1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯(1 + x)^n = 1 + nx + \frac{n(n – 1)}{2!}x^2 + \frac{n(n – 1)(n – 2)}{3!}x^3 + \cdots(1+x)n=1+nx+2!n(n−1)​x2+3!n(n−1)(n−2)​x3+⋯

This is an infinite series (convergent for ∣x∣<1|x| < 1∣x∣<1).

🔹 Example 1 — Fractional Power

Expand (1+x)1/2(1 + x)^{1/2}(1+x)1/2 up to four terms. (1+x)1/2=1+12x+(1/2)(−1/2)2!x2+(1/2)(−1/2)(−3/2)3!x3+⋯(1 + x)^{1/2} = 1 + \frac{1}{2}x + \frac{(1/2)(-1/2)}{2!}x^2 + \frac{(1/2)(-1/2)(-3/2)}{3!}x^3 + \cdots(1+x)1/2=1+21​x+2!(1/2)(−1/2)​x2+3!(1/2)(−1/2)(−3/2)​x3+⋯

Simplify: (1+x)1/2=1+x2−x28+x316−⋯(1 + x)^{1/2} = 1 + \frac{x}{2} – \frac{x^2}{8} + \frac{x^3}{16} – \cdots(1+x)1/2=1+2x​−8×2​+16×3​−⋯

✅ This gives the power series for 1+x\sqrt{1 + x}1+x​.

🔹 Example 2 — Negative Power

Expand (1+x)−1(1 + x)^{-1}(1+x)−1. (1+x)−1=1+(−1)x+(−1)(−2)x22!+(−1)(−2)(−3)x33!+⋯(1 + x)^{-1} = 1 + (-1)x + (-1)(-2)\frac{x^2}{2!} + (-1)(-2)(-3)\frac{x^3}{3!} + \cdots(1+x)−1=1+(−1)x+(−1)(−2)2!x2​+(−1)(−2)(−3)3!x3​+⋯

Simplify: (1+x)−1=1−x+x2−x3+x4−⋯(1 + x)^{-1} = 1 – x + x^2 – x^3 + x^4 – \cdots(1+x)−1=1−x+x2−x3+x4−⋯

✅ This is the geometric series with ratio −x-x−x.

🔹 Example 3 — Negative Fractional Power

Expand (1+x)−1/2(1 + x)^{-1/2}(1+x)−1/2 up to 4 terms. (1+x)−1/2=1+(−12)x+(−12)(−32)2!x2−(−12)(−32)(−52)3!x3+⋯(1 + x)^{-1/2} = 1 + \left(-\frac{1}{2}\right)x + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}x^2 – \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}x^3 + \cdots(1+x)−1/2=1+(−21​)x+2!(−21​)(−23​)​x2−3!(−21​)(−23​)(−25​)​x3+⋯

Simplify: (1+x)−1/2=1−x2+3×28−5×316+⋯(1 + x)^{-1/2} = 1 – \frac{x}{2} + \frac{3x^2}{8} – \frac{5x^3}{16} + \cdots(1+x)−1/2=1−2x​+83×2​−165×3​+⋯

✅ This series represents 11+x\frac{1}{\sqrt{1 + x}}1+x​1​.

💡 4️⃣ Important Observations

🔹 Why ∣x∣<1|x| < 1∣x∣<1?

Because as r→∞r \to \inftyr→∞, the terms xrx^rxr get smaller only if ∣x∣<1|x| < 1∣x∣<1.
If ∣x∣≥1|x| ≥ 1∣x∣≥1, the infinite series diverges.

🔹 Shortcut Expansions to Remember

| Function | Expansion (|x| < 1) |
|———–|————-|
| (1+x)−1(1 + x)^{-1}(1+x)−1 | 1−x+x2−x3+⋯1 – x + x^2 – x^3 + \cdots1−x+x2−x3+⋯ |
| (1+x)−2(1 + x)^{-2}(1+x)−2 | 1−2x+3×2−4×3+⋯1 – 2x + 3x^2 – 4x^3 + \cdots1−2x+3×2−4×3+⋯ |
| (1+x)1/2(1 + x)^{1/2}(1+x)1/2 | 1+x2−x28+x316−⋯1 + \frac{x}{2} – \frac{x^2}{8} + \frac{x^3}{16} – \cdots1+2x​−8×2​+16×3​−⋯ |
| (1+x)−1/2(1 + x)^{-1/2}(1+x)−1/2 | 1−x2+3×28−5×316+⋯1 – \frac{x}{2} + \frac{3x^2}{8} – \frac{5x^3}{16} + \cdots1−2x​+83×2​−165×3​+⋯ |

These appear frequently in IB, A Level, and STEP questions.

🔹 Quick IB-Style Application

Find the first three terms of 14−x\frac{1}{\sqrt{4 – x}}4−x​1​.

Rewrite: 14−x=12(1−x4)−1/2\frac{1}{\sqrt{4 – x}} = \frac{1}{2}\left(1 – \frac{x}{4}\right)^{-1/2}4−x​1​=21​(1−4x​)−1/2

Let u=−x4u = -\frac{x}{4}u=−4x​.
Then: (1+u)−1/2=1−12u+38u2+⋯(1 + u)^{-1/2} = 1 – \frac{1}{2}u + \frac{3}{8}u^2 + \cdots(1+u)−1/2=1−21​u+83​u2+⋯ 14−x=12(1+x8+3×2128+⋯ )\frac{1}{\sqrt{4 – x}} = \frac{1}{2}\left(1 + \frac{x}{8} + \frac{3x^2}{128} + \cdots\right)4−x​1​=21​(1+8x​+1283×2​+⋯)

✅ Result: 12+x16+3×2256+⋯\frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256} + \cdots21​+16x​+2563×2​+⋯

🚀 5️⃣ Why This Generalization Matters

This extension allows us to:

• Expand non-integer powers in physics & calculus,
• Derive Taylor series easily,
• Handle rational exponents in approximations,
• Model real-world nonlinear phenomena mathematically.

🌟 Summary Table

📘 Learn Beyond Formula

At Math By Rishabh, we don’t just teach expansions — we teach why they work.

In the Mathematics Elevate Mentorship Program, you’ll:
✅ Derive generalized binomial expansions from first principles,
✅ Apply them to STEP/MAT-level problems,
✅ Connect algebraic series to calculus and approximation.

🚀 Learn to expand beyond limits — literally.
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