Binomial Theorem for Positive Integer
Introduction
The Binomial Theorem provides a formula for expanding $(a + b)^n$ where $n$ is a positive integer. This theorem is fundamental in algebra and has numerous applications in JEE problems.
Binomial Theorem Statement
Binomial Theorem: For any positive integer $n$,
$$\boxed{(a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n}$$Or equivalently,
$$\boxed{(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r}$$
where $\binom{n}{r} = \frac{n!}{r!(n-r)!} = {}^nC_r$
Key Properties
1. Number of Terms
- The expansion of $(a + b)^n$ has exactly $(n + 1)$ terms
2. Powers in Each Term
$$\boxed{\text{Sum of powers of } a \text{ and } b \text{ in each term} = n}$$
3. Binomial Coefficients
The coefficients $\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n}$ are called binomial coefficients.
Important Forms
Form 1: $(1 + x)^n$
$$\boxed{(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots + x^n}$$Form 2: $(1 - x)^n$
$$\boxed{(1 - x)^n = 1 - nx + \frac{n(n-1)}{2!}x^2 - \frac{n(n-1)(n-2)}{3!}x^3 + \cdots + (-1)^n x^n}$$Form 3: $(x + a)^n$
$$\boxed{(x + a)^n = x^n + nx^{n-1}a + \frac{n(n-1)}{2!}x^{n-2}a^2 + \cdots + a^n}$$Memory Tricks
🎯 Pascal’s Triangle Method
Interactive Demo: Visualize Pascal’s Triangle
Explore binomial coefficients through Pascal’s Triangle visualization.
Remember the coefficients using Pascal’s Triangle:
n=0: 1
n=1: 1 1
n=2: 1 2 1
n=3: 1 3 3 1
n=4: 1 4 6 4 1
n=5: 1 5 10 10 5 1
Each number is the sum of the two numbers directly above it.
🎯 Power Pattern
“Powers go opposite ways”
- Power of first term: decreases from $n$ to $0$
- Power of second term: increases from $0$ to $n$
- Remember: “Down-Up” pattern
🎯 Coefficient Formula
“nCr = n Choose r”
- Start with $n$ on top
- $r$ items in the denominator on right
- $(n-r)$ items on left
- Example: $\binom{5}{2} = \frac{5 \times 4}{2 \times 1}$
Common Mistakes to Avoid
❌ Mistake 1: Forgetting the Factorial Division
Wrong: $(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$ ✗ (coefficients are wrong)
Correct: Use $\binom{3}{r}$ for coefficients:
- $\binom{3}{0} = 1$, $\binom{3}{1} = 3$, $\binom{3}{2} = 3$, $\binom{3}{3} = 1$
❌ Mistake 2: Sign Errors in $(a - b)^n$
Wrong: $(x - 2)^3 = x^3 - 3x^2(2) - 3x(4) - 8$ ✗
Correct: $(x - 2)^3 = x^3 + 3x^2(-2) + 3x(-2)^2 + (-2)^3$ $= x^3 - 6x^2 + 12x - 8$ ✓
Tip: Replace $b$ with $(-b)$ and apply the theorem.
❌ Mistake 3: Power Sum Not Equal to $n$
Wrong: In $(x + y)^5$, term $x^3y^3$ exists ✗
Correct: Sum of powers must be 5, so $x^3y^3$ (sum = 6) doesn’t exist ✓
❌ Mistake 4: Incorrect Coefficient Calculation
Wrong: $\binom{6}{2} = 6 \times 2 = 12$ ✗
Correct: $\binom{6}{2} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$ ✓
Solved Examples
Example 1: Basic Expansion (JEE Main Level)
Expand $(2x + 3y)^4$ using binomial theorem.
Solution: Using $(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$
Here $a = 2x$, $b = 3y$, $n = 4$
$$\begin{align} (2x + 3y)^4 &= \binom{4}{0}(2x)^4 + \binom{4}{1}(2x)^3(3y) + \binom{4}{2}(2x)^2(3y)^2 \\ &\quad + \binom{4}{3}(2x)(3y)^3 + \binom{4}{4}(3y)^4 \\ &= 1(16x^4) + 4(8x^3)(3y) + 6(4x^2)(9y^2) \\ &\quad + 4(2x)(27y^3) + 1(81y^4) \\ &= 16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4 \end{align}$$Example 2: Finding Specific Term (JEE Main Level)
Find the coefficient of $x^5$ in the expansion of $(2 + 3x)^7$.
Solution: General term: $T_{r+1} = \binom{7}{r}(2)^{7-r}(3x)^r = \binom{7}{r}2^{7-r}3^r x^r$
For term containing $x^5$: $r = 5$
$$\begin{align} T_6 &= \binom{7}{5}(2)^{2}(3)^5 x^5 \\ &= 21 \times 4 \times 243 \times x^5 \\ &= 20412x^5 \end{align}$$Coefficient of $x^5$ = 20412
Example 3: Numerical Calculation (JEE Advanced Level)
Using binomial theorem, find the value of $(1.01)^5$ correct to 4 decimal places.
Solution: $(1.01)^5 = (1 + 0.01)^5$
$$\begin{align} (1 + x)^5 &= 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5 \end{align}$$Substituting $x = 0.01$:
$$\begin{align} (1.01)^5 &= 1 + 5(0.01) + 10(0.01)^2 + 10(0.01)^3 + 5(0.01)^4 + (0.01)^5 \\ &= 1 + 0.05 + 0.001 + 0.00001 + 0.0000005 + 0.00000001 \\ &= 1.0510 \text{ (correct to 4 decimal places)} \end{align}$$Practice Problems
Level 1: JEE Main Basics
Problem 1.1: Expand $(x + 2)^5$ using binomial theorem.
Solution
$$\begin{align} (x + 2)^5 &= x^5 + 5x^4(2) + 10x^3(4) + 10x^2(8) + 5x(16) + 32 \\ &= x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32 \end{align}$$Problem 1.2: Find the number of terms in the expansion of $(3x - 2y)^{10}$.
Solution
Number of terms = $n + 1 = 10 + 1 = 11$
Problem 1.3: Write the general term in the expansion of $(a + b)^{12}$.
Solution
General term: $T_{r+1} = \binom{12}{r}a^{12-r}b^r$ where $r = 0, 1, 2, \ldots, 12$
Level 2: JEE Main Advanced
Problem 2.1: Find the coefficient of $x^6$ in the expansion of $(1 + 2x)^{10}$.
Solution
$T_{r+1} = \binom{10}{r}(2x)^r = \binom{10}{r}2^r x^r$
For $x^6$: $r = 6$
Coefficient $= \binom{10}{6}2^6 = 210 \times 64 = 13440$
Problem 2.2: If the coefficients of $x^7$ and $x^8$ in $(2 + x/3)^n$ are equal, find $n$.
Solution
Coefficient of $x^7$: $\binom{n}{7}2^{n-7}\left(\frac{1}{3}\right)^7$
Coefficient of $x^8$: $\binom{n}{8}2^{n-8}\left(\frac{1}{3}\right)^8$
Setting them equal:
$$\binom{n}{7}2^{n-7} \cdot 3^{-7} = \binom{n}{8}2^{n-8} \cdot 3^{-8}$$ $$\frac{n!}{7!(n-7)!} \cdot 2 = \frac{n!}{8!(n-8)!} \cdot 3$$ $$\frac{8}{n-7} \cdot 2 = 3$$ $$16 = 3(n - 7)$$ $$n = \frac{16}{3} + 7 = \frac{37}{3}$$Wait, this should give integer. Let me recalculate:
$$\binom{n}{7} \cdot 2 \cdot 3 = \binom{n}{8}$$ $$\frac{n!}{7!(n-7)!} \cdot 6 = \frac{n!}{8!(n-8)!}$$ $$\frac{8!}{7!} \cdot \frac{(n-8)!}{(n-7)!} = \frac{1}{6}$$ $$8 \cdot \frac{1}{n-7} = \frac{1}{6}$$ $$n - 7 = 48$$, so $n = 55$
Problem 2.3: Calculate $(0.99)^6$ using binomial theorem (up to 4 decimal places).
Solution
$(0.99)^6 = (1 - 0.01)^6$
$= 1 - 6(0.01) + 15(0.01)^2 - 20(0.01)^3 + \ldots$
$= 1 - 0.06 + 0.0015 - 0.00002 + \ldots$
$= 0.9415$ (to 4 decimal places)
Level 3: JEE Advanced
Problem 3.1: Find the term independent of $x$ in the expansion of $\left(2x^2 - \frac{1}{x}\right)^{12}$.
Solution
General term: $T_{r+1} = \binom{12}{r}(2x^2)^{12-r}\left(-\frac{1}{x}\right)^r$
$= \binom{12}{r}2^{12-r}(-1)^r x^{24-2r-r}$
$= \binom{12}{r}2^{12-r}(-1)^r x^{24-3r}$
For term independent of $x$: $24 - 3r = 0 \Rightarrow r = 8$
$T_9 = \binom{12}{8}2^4(-1)^8 = 495 \times 16 \times 1 = 7920$
Problem 3.2: If the sum of coefficients in the expansion of $(x + y)^n$ is 4096, find the greatest coefficient.
Solution
Sum of coefficients = Put $x = y = 1$: $(1 + 1)^n = 2^n = 4096$
So $n = 12$
For $(x + y)^{12}$, greatest coefficient is the middle one: $\binom{12}{6} = 924$
Problem 3.3: Prove that $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = 2^n$.
Solution
Consider $(1 + 1)^n = \sum_{r=0}^{n}\binom{n}{r}(1)^{n-r}(1)^r = \sum_{r=0}^{n}\binom{n}{r}$
Therefore, $\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n$
Cross-References
- Sequences and Series: Binomial coefficients form interesting sequences → Sequences and Series
- Differentiation: Binomial theorem helps in differential calculus → Differentiation
- General Term: Learn about finding specific terms → General Term
- Binomial Coefficients: Properties and identities → Binomial Coefficients
Quick Revision Points
- $(a + b)^n$ has $(n + 1)$ terms
- General term: $T_{r+1} = \binom{n}{r}a^{n-r}b^r$
- Sum of powers in each term = $n$
- For $(1 + x)^n$: Put $a = 1, b = x$
- For $(1 - x)^n$: Replace $x$ with $-x$
- Pascal’s Triangle gives binomial coefficients
- Sum of all coefficients: Put $a = b = 1$
Last updated: October 15, 2025