Applications of Binomial Theorem in Approximations
Introduction
The Binomial Theorem is not just a theoretical tool—it has powerful applications in numerical approximations, error estimation, series expansions, and solving JEE problems efficiently.
Binomial Approximation Formula
For small values of $x$ (typically $|x| < 1$):
$$\boxed{(1 + x)^n \approx 1 + nx}$$
Interactive Demo: Visualize Binomial Approximations
See how binomial expansion terms relate to Pascal’s Triangle.
This is the first-order approximation (keeping only first two terms).
More Accurate Approximation
$$\boxed{(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2}$$
This is the second-order approximation (keeping first three terms).
General Form
$$\boxed{(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots}$$
Valid for: $|x| < 1$ and any real $n$ (positive, negative, or fraction)
Standard Approximations
Type 1: Computing Powers Near 1
For $(1 + x)^n$ where $x$ is small:
$$\boxed{(1 + x)^n \approx 1 + nx}$$Examples:
- $(1.01)^5 \approx 1 + 5(0.01) = 1.05$
- $(0.99)^{10} = (1 - 0.01)^{10} \approx 1 + 10(-0.01) = 0.90$
Type 2: Square Roots
For $\sqrt{1 + x}$ where $x$ is small:
$$\boxed{\sqrt{1 + x} = (1 + x)^{1/2} \approx 1 + \frac{x}{2} - \frac{x^2}{8}}$$First order: $\sqrt{1 + x} \approx 1 + \frac{x}{2}$
Examples:
- $\sqrt{1.04} = \sqrt{1 + 0.04} \approx 1 + \frac{0.04}{2} = 1.02$
- $\sqrt{0.96} = \sqrt{1 - 0.04} \approx 1 - \frac{0.04}{2} = 0.98$
Type 3: Cube Roots
For $\sqrt[3]{1 + x}$ where $x$ is small:
$$\boxed{\sqrt[3]{1 + x} = (1 + x)^{1/3} \approx 1 + \frac{x}{3} - \frac{x^2}{9}}$$First order: $\sqrt[3]{1 + x} \approx 1 + \frac{x}{3}$
Examples:
- $\sqrt[3]{1.06} \approx 1 + \frac{0.06}{3} = 1.02$
- $\sqrt[3]{0.97} \approx 1 - \frac{0.03}{3} = 0.99$
Type 4: Reciprocals
For $\frac{1}{1 + x}$ where $x$ is small:
$$\boxed{\frac{1}{1 + x} = (1 + x)^{-1} \approx 1 - x + x^2 - x^3 + \cdots}$$First order: $\frac{1}{1 + x} \approx 1 - x$
Examples:
- $\frac{1}{1.02} \approx 1 - 0.02 = 0.98$
- $\frac{1}{0.98} = \frac{1}{1 - 0.02} \approx 1 + 0.02 = 1.02$
Conversion to Standard Form
To use binomial approximation, convert expressions to the form $(1 + x)^n$:
Method 1: Factor Out
Example: Approximate $(2.03)^5$
$$\begin{align} (2.03)^5 &= (2 + 0.03)^5 \\ &= 2^5\left(1 + \frac{0.03}{2}\right)^5 \\ &= 32(1 + 0.015)^5 \\ &\approx 32[1 + 5(0.015)] \\ &= 32(1.075) \\ &= 34.4 \end{align}$$Method 2: Rewrite Form
Example: Approximate $\sqrt{99}$
$$\begin{align} \sqrt{99} &= \sqrt{100 - 1} \\ &= 10\sqrt{1 - \frac{1}{100}} \\ &= 10(1 - 0.01)^{1/2} \\ &\approx 10\left[1 + \frac{1}{2}(-0.01)\right] \\ &= 10(1 - 0.005) \\ &= 9.95 \end{align}$$Error Estimation
The error in approximation depends on terms we neglected.
Relative Error
If we use $(1 + x)^n \approx 1 + nx$, the relative error is approximately:
$$\boxed{\text{Relative Error} \approx \frac{n(n-1)x^2}{2}}$$
Example: For $(1.01)^{10}$ using $(1 + x)^n \approx 1 + nx$:
Error $\approx \frac{10 \times 9 \times (0.01)^2}{2} = 0.0045 = 0.45\%$
Applications in Different Areas
Application 1: Finding Greatest Term
To find numerically greatest term in $(a + b)^n$:
The greatest term is $T_{r+1}$ where:
$$\boxed{r = \left[\frac{(n+1)|b|}{|a| + |b|}\right]}$$where $[x]$ denotes greatest integer function.
Alternative Method: $T_{r+1} > T_r$ when:
$$\frac{(n - r + 1)|b|}{r|a|} > 1$$Application 2: Divisibility Problems
Check if $(a + b)^n$ is divisible by some number:
Use binomial expansion and modular arithmetic.
Example: Is $7^{100}$ divisible by 25?
$$7^{100} = (1 + 6)^{100} = 1 + 100(6) + \binom{100}{2}(6)^2 + \cdots$$$= 1 + 600 + 4950(36) + \cdots$
First two terms: $1 + 600 = 601$ (divisible by 25? No, but continue…)
All terms after second contain $6^2 = 36$, and with large binomial coefficients, we can analyze modulo 25.
Application 3: Proving Inequalities
Using binomial theorem to prove inequalities:
Example: Prove $2^n > 1 + n$ for $n \geq 2$.
Proof:
$$2^n = (1 + 1)^n = 1 + n + \binom{n}{2} + \binom{n}{3} + \cdots$$Since all binomial coefficients are positive for $n \geq 2$:
$$2^n > 1 + n$$Application 4: Sum of Series
Finding sums using binomial identities:
Example: Find $1 + 2x + 3x^2 + 4x^3 + \cdots + (n+1)x^n$
This is related to differentiating $(1 + x)^{n+1}$.
Memory Tricks
🎯 Small x Approximation
“One plus n times x”: $(1 + x)^n \approx 1 + nx$ for small $x$
Mnemonic: “When Small, Keep Two”
- When $x$ is small, keep (first) two terms
🎯 Root Approximations
Square root: “Half the difference” $\sqrt{1 + x} \approx 1 + \frac{x}{2}$
Cube root: “Third the difference” $\sqrt[3]{1 + x} \approx 1 + \frac{x}{3}$
$n$-th root: “$\frac{1}{n}$ the difference” $(1 + x)^{1/n} \approx 1 + \frac{x}{n}$
🎯 Conversion Trick
“Factor and Convert”:
- Factor out to make base close to 1
- Convert to $(1 + \text{small number})^n$
- Apply approximation
Common Mistakes to Avoid
❌ Mistake 1: Using Approximation for Large x
Wrong: $(1 + 0.5)^{10} \approx 1 + 10(0.5) = 6$ ✗
Correct: $x = 0.5$ is not small enough for first-order approximation. Use more terms or calculate exactly: $(1.5)^{10} = 57.665$ ✓
Rule: Use approximation only when $|x| \ll 1$ (typically $|x| < 0.1$)
❌ Mistake 2: Forgetting Sign in $(1 - x)^n$
Wrong: $(0.98)^5 = (1 - 0.02)^5 \approx 1 + 5(0.02)$ ✗
Correct: $(1 - 0.02)^5 \approx 1 + 5(-0.02) = 1 - 0.1 = 0.9$ ✓
❌ Mistake 3: Incorrect Factoring
Wrong: $(3.1)^4 = 3^4(1 + 0.1)^4$ ✗
Correct: $(3.1)^4 = (3 + 0.1)^4 = 3^4\left(1 + \frac{0.1}{3}\right)^4$ ✓
❌ Mistake 4: Wrong Root Formula
Wrong: $\sqrt{1 + x} \approx 1 + x$ ✗
Correct: $\sqrt{1 + x} = (1 + x)^{1/2} \approx 1 + \frac{x}{2}$ ✓
Solved Examples
Example 1: Power Approximation (JEE Main)
Find the approximate value of $(1.02)^{10}$ using binomial theorem.
Solution: $(1.02)^{10} = (1 + 0.02)^{10}$
Using $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots$
First-order approximation:
$$(1.02)^{10} \approx 1 + 10(0.02) = 1.2$$Second-order approximation:
$$\begin{align} (1.02)^{10} &\approx 1 + 10(0.02) + \frac{10 \times 9}{2}(0.02)^2 \\ &= 1 + 0.2 + 45(0.0004) \\ &= 1 + 0.2 + 0.018 \\ &= 1.218 \end{align}$$Exact value: $(1.02)^{10} = 1.21899...$
Example 2: Square Root (JEE Main)
Calculate $\sqrt{26}$ approximately.
Solution:
$$\begin{align} \sqrt{26} &= \sqrt{25 + 1} \\ &= 5\sqrt{1 + \frac{1}{25}} \\ &= 5(1 + 0.04)^{1/2} \\ &\approx 5\left[1 + \frac{1}{2}(0.04)\right] \\ &= 5(1 + 0.02) \\ &= 5.1 \end{align}$$Exact value: $\sqrt{26} = 5.0990...$
Error: $5.1 - 5.099 = 0.001$ (very small!)
Example 3: Reciprocal Approximation (JEE Advanced)
Find the value of $\frac{1}{(2.002)^3}$ up to 4 decimal places.
Solution:
$$\begin{align} \frac{1}{(2.002)^3} &= \frac{1}{2^3(1 + 0.001)^3} \\ &= \frac{1}{8} \cdot \frac{1}{(1 + 0.001)^3} \\ &= \frac{1}{8}(1 + 0.001)^{-3} \end{align}$$Using $(1 + x)^{-3} = 1 - 3x + 6x^2 - 10x^3 + \cdots$:
$$\begin{align} &\approx \frac{1}{8}[1 - 3(0.001) + 6(0.001)^2] \\ &= \frac{1}{8}[1 - 0.003 + 0.000006] \\ &= \frac{1}{8}(0.997006) \\ &= 0.1246 \text{ (to 4 decimal places)} \end{align}$$Example 4: Cube Root (JEE Advanced)
Evaluate $(127)^{1/3}$ approximately.
Solution:
$$\begin{align} (127)^{1/3} &= (125 + 2)^{1/3} \\ &= 125^{1/3}\left(1 + \frac{2}{125}\right)^{1/3} \\ &= 5(1 + 0.016)^{1/3} \\ &\approx 5\left[1 + \frac{1}{3}(0.016)\right] \\ &= 5(1 + 0.00533) \\ &= 5.0267 \end{align}$$Exact value: $(127)^{1/3} = 5.0265...$
Example 5: Mixed Application (JEE Advanced)
If $(1 + x)^n = C_0 + C_1x + C_2x^2 + \cdots + C_nx^n$, find approximately the value of $C_0 + C_1/2 + C_2/4 + C_3/8 + \cdots + C_n/2^n$ when $n = 10$.
Solution: We need: $C_0 + \frac{C_1}{2} + \frac{C_2}{4} + \cdots + \frac{C_{10}}{2^{10}}$
This is $(1 + x)^n$ evaluated at $x = 1/2$:
$$(1 + 1/2)^{10} = (1.5)^{10}$$Converting: $(1.5)^{10} = \left(\frac{3}{2}\right)^{10} = \frac{3^{10}}{2^{10}}$
$= \frac{59049}{1024} = 57.665$
Or using approximation: $1.5 = 1 + 0.5$
But $x = 0.5$ is too large for simple approximation. Better to calculate exactly or use logarithms.
Practice Problems
Level 1: JEE Main Basics
Problem 1.1: Find $(0.99)^5$ approximately.
Solution
$(0.99)^5 = (1 - 0.01)^5 \approx 1 + 5(-0.01) = 1 - 0.05 = 0.95$
More accurate: $1 - 5(0.01) + 10(0.01)^2 = 1 - 0.05 + 0.001 = 0.951$
Exact: $(0.99)^5 = 0.9510...$
Problem 1.2: Approximate $\sqrt{37}$.
Solution
$\sqrt{37} = \sqrt{36 + 1} = 6\sqrt{1 + 1/36} = 6(1 + 1/36)^{1/2}$
$\approx 6\left[1 + \frac{1}{2} \cdot \frac{1}{36}\right] = 6\left[1 + \frac{1}{72}\right] = 6.0833$
Exact: $\sqrt{37} = 6.0828...$
Problem 1.3: Find $(1.05)^3$ using binomial expansion.
Solution
$(1.05)^3 = (1 + 0.05)^3$
$= 1 + 3(0.05) + 3(0.05)^2 + (0.05)^3$
$= 1 + 0.15 + 3(0.0025) + 0.000125$
$= 1 + 0.15 + 0.0075 + 0.000125$
$= 1.157625$
Level 2: JEE Main Advanced
Problem 2.1: Find the approximate value of $(0.998)^{10}$ correct to four decimal places.
Solution
$(0.998)^{10} = (1 - 0.002)^{10}$
$\approx 1 + 10(-0.002) + \frac{10 \times 9}{2}(0.002)^2 + \frac{10 \times 9 \times 8}{6}(0.002)^3$
$= 1 - 0.02 + 45(0.000004) + 120(0.000000008)$
$= 1 - 0.02 + 0.00018 + 0.00000096$
$= 0.9802$ (to 4 decimal places)
Problem 2.2: Using binomial theorem, prove that $n! > \left(\frac{n}{e}\right)^n$ for $n \geq 1$.
Solution
Consider $(1 + 1/n)^n = 1 + n \cdot \frac{1}{n} + \frac{n(n-1)}{2!} \cdot \frac{1}{n^2} + \cdots$
$= 1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \cdots$
As $n \to \infty$: $(1 + 1/n)^n \to e$
For finite $n$: $(1 + 1/n)^n < e$, so $n < e^n \cdot n^n / n^n$…
(This requires Stirling’s approximation for complete proof)
Problem 2.3: Find $\sqrt[4]{17}$ approximately.
Solution
$\sqrt[4]{17} = (16 + 1)^{1/4} = 2(1 + 1/16)^{1/4}$
$\approx 2\left[1 + \frac{1}{4} \cdot \frac{1}{16}\right] = 2\left[1 + \frac{1}{64}\right] = 2.03125$
Exact: $\sqrt[4]{17} = 2.0305...$
Level 3: JEE Advanced
Problem 3.1: Show that for large $n$, $\left(1 + \frac{1}{n}\right)^n \approx e\left(1 - \frac{1}{2n}\right)$ where $e = 2.718...$
Solution
Using binomial expansion and taking logarithm:
$\ln\left[\left(1 + \frac{1}{n}\right)^n\right] = n\ln\left(1 + \frac{1}{n}\right)$
For small $x$: $\ln(1 + x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$
$n\ln\left(1 + \frac{1}{n}\right) \approx n\left[\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \cdots\right]$
$= 1 - \frac{1}{2n} + \frac{1}{3n^2} - \cdots \approx 1 - \frac{1}{2n}$ for large $n$
Therefore: $\left(1 + \frac{1}{n}\right)^n \approx e^{1-1/(2n)} = e \cdot e^{-1/(2n)} \approx e\left(1 - \frac{1}{2n}\right)$
Problem 3.2: If $x$ is so small that $x^3$ and higher powers can be neglected, show that:
$$\frac{(1-x)^{3/2}}{(1+x)^{1/2}} \approx 1 - 2x$$Solution
$(1-x)^{3/2} \approx 1 + \frac{3}{2}(-x) + \frac{3/2 \cdot 1/2}{2}(-x)^2$
$= 1 - \frac{3x}{2} + \frac{3x^2}{8}$
Neglecting $x^2$: $(1-x)^{3/2} \approx 1 - \frac{3x}{2}$
$(1+x)^{-1/2} \approx 1 + \left(-\frac{1}{2}\right)x = 1 - \frac{x}{2}$
Therefore: $\frac{(1-x)^{3/2}}{(1+x)^{1/2}} \approx \left(1 - \frac{3x}{2}\right)\left(1 - \frac{x}{2}\right)$
$= 1 - \frac{3x}{2} - \frac{x}{2} + \frac{3x^2}{4}$
Neglecting $x^2$: $\approx 1 - 2x$
Problem 3.3: Find the percentage error in taking $(0.997)^{1/3}$ as $1 - \frac{0.003}{3}$.
Solution
Approximation: $(0.997)^{1/3} \approx 1 - \frac{0.003}{3} = 1 - 0.001 = 0.999$
Better approximation: $(0.997)^{1/3} = (1 - 0.003)^{1/3}$
$\approx 1 + \frac{1}{3}(-0.003) + \frac{(1/3)(-2/3)}{2}(-0.003)^2$
$= 1 - 0.001 - \frac{1}{9}(0.000009)$
$= 1 - 0.001 - 0.000001 = 0.999999$
Percentage error $= \frac{0.999999 - 0.999}{0.999999} \times 100 \approx 0.0001\%$
The error is negligible!
Quick Reference Table
| Expression | Approximation (small $x$) |
|---|---|
| $(1 + x)^n$ | $1 + nx$ |
| $(1 - x)^n$ | $1 - nx$ |
| $\sqrt{1 + x}$ | $1 + \frac{x}{2}$ |
| $\sqrt[3]{1 + x}$ | $1 + \frac{x}{3}$ |
| $\frac{1}{1 + x}$ | $1 - x$ |
| $\frac{1}{\sqrt{1 + x}}$ | $1 - \frac{x}{2}$ |
| $\frac{1}{(1 + x)^2}$ | $1 - 2x$ |
Cross-References
- Binomial Expansion: Core concepts → Binomial Expansion
- Calculus Applications: Taylor series connection → Differentiation
- Limits: Approximations in limits → Limits
Quick Revision Points
- Use binomial approximation only for $|x| \ll 1$
- First-order: Keep two terms
- Second-order: Keep three terms
- Always convert to $(1 + x)^n$ form
- Check sign carefully for $(1 - x)^n$
- Root formulas: power is $1/n$
Last updated: October 25, 2025