NCERT Chapter 7

Binomial Theorem

Expand any power $(a + b)^n$ effortlessly: from Pascal's triangle to the general term, middle terms, and divisibility proofs.

~40 min read 4–6 marks in exam kemh107.pdf

What you will learn in this chapter

  1. Introduction & The Geometry of Pascal's Triangle
  2. The Binomial Theorem Statement & Core Formula
  3. Important Special Expansions & Coefficient Properties
  4. General Term Formula ($T_{r+1}$)
  5. Finding the Middle Term(s) (Even vs Odd $n$)
  6. Term Independent of $x$ & Divisibility Proofs
  7. Worked Examples — Easy to Hard
  8. NCERT Key Exercise Questions Solved
  9. Common Exam Traps & Pitfalls
  10. Quick Self-Check Quiz

1. Pascal's Triangle & Motivation

Why We Need the Binomial Theorem

We all know $(a + b)^2 = a^2 + 2ab + b^2$ and $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. But what if an exam asks you to calculate $(a + b)^{10}$ or evaluate $(99)^5$? Multiplying brackets 10 times by hand would take hours. The Binomial Theorem gives an exact algebraic shortcut for any power $n$.

Pascal's Triangle Pattern

Arranging the coefficients of $(a+b)^n$ for $n = 0, 1, 2, 3, 4, 5$ produces a triangle where each number is the sum of the two numbers directly above it:

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

Notice that row $n$ consists of combinations: $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$.

Because $^nC_r + ^nC_{r-1} = ^{n+1}C_r$, every entry is the sum of the two above it!

2. The Binomial Theorem Statement

Theorem: Binomial Theorem for Any Positive Integer $n$

$$(a + b)^n = ^nC_0 a^n + ^nC_1 a^{n-1}b + ^nC_2 a^{n-2}b^2 + \dots + ^nC_r a^{n-r}b^r + \dots + ^nC_n b^n$$

In summation sigma notation:

$$(a + b)^n = \sum_{k=0}^n {^nC_k} a^{n-k} b^k$$

3 Golden Observations You Must Remember

  1. Number of Terms: The expansion of $(a + b)^n$ contains exactly $\mathbf{n + 1}$ terms. (E.g. $(x+y)^7$ has $7 + 1 = 8$ terms).
  2. Conservation of Powers: In every single term, the power of $a$ decreases from $n$ to $0$, the power of $b$ increases from $0$ to $n$, and the sum of exponents is ALWAYS $n$: $(n - r) + r = n$.
  3. Symmetry of Coefficients: Coefficients of terms equidistant from the beginning and end are equal, because $^nC_r = ^nC_{n - r}$.

3. Special Expansions & Coefficient Identities

Expansion of $(a - b)^n$

Replace $b$ with $(-b)$. The signs alternate strictly ($+$, $-$, $+$, $-$, ...):

$$(a - b)^n = ^nC_0 a^n - ^nC_1 a^{n-1}b + ^nC_2 a^{n-2}b^2 - \dots + (-1)^n ^nC_n b^n$$

Expansion of $(1 + x)^n$

Set $a = 1$ and $b = x$:

$$(1 + x)^n = 1 + ^nC_1 x + ^nC_2 x^2 + ^nC_3 x^3 + \dots + ^nC_n x^n$$

Sum of All Binomial Coefficients

Plug $x = 1$ into $(1 + x)^n$:

$$^nC_0 + ^nC_1 + ^nC_2 + \dots + ^nC_n = (1 + 1)^n = \mathbf{2^n}$$

Sum of Odd = Sum of Even Coefficients

Plug $x = -1$ into $(1 + x)^n$:

$$^nC_0 + ^nC_2 + ^nC_4 + \dots = ^nC_1 + ^nC_3 + ^nC_5 + \dots = \mathbf{2^{n-1}}$$

4. The General Term Formula ($T_{r+1}$)

The General Term Formula

The $(r + 1)$-th term in the expansion of $(a + b)^n$ is denoted by $T_{r+1}$:

$$\mathbf{T_{r+1} = ^nC_r \cdot a^{n - r} \cdot b^r}$$

Crucial Off-By-One Warning!

Remember that the index $r$ is always ONE LESS than the term number:

5. Finding Middle Terms

Case 1: When $n$ is EVEN

The total number of terms $(n + 1)$ is ODD. Therefore, there is EXACTLY ONE middle term:

$$\text{Middle Term} = T_{\frac{n}{2} + 1}$$

Example: For $(x + y)^8$, $n = 8$ is even. Middle term is $T_{8/2 + 1} = T_5$.

Case 2: When $n$ is ODD

The total number of terms $(n + 1)$ is EVEN. Therefore, there are TWO middle terms:

$$\text{Middle Terms} = T_{\frac{n + 1}{2}} \quad \text{and} \quad T_{\frac{n + 3}{2}}$$

Example: For $(x + y)^7$, $n = 7$ is odd. Middle terms are $T_4$ and $T_5$.

6. Term Independent of $x$ & Divisibility Proofs

1. Term Independent of $x$ (Constant Term)

To find the term containing no $x$ (i.e. $x^0$):

  1. Write out the general term $T_{r+1} = ^nC_r a^{n-r} b^r$.
  2. Collect all powers of $x$ into a single exponent: $x^{f(r)}$.
  3. Set the exponent $f(r) = 0$ and solve for integer $r$.
  4. Substitute that $r$ back into $T_{r+1}$ to obtain the numerical constant value.

2. Divisibility Questions

To prove an expression like $9^{n+1} - 8n - 9$ is divisible by $64$ ($= 8^2$):

Write base $9$ as $(1 + 8)$ and expand using Binomial Theorem:

$$9^{n+1} = (1 + 8)^{n+1} = ^{n+1}C_0 + ^{n+1}C_1(8) + ^{n+1}C_2(8)^2 + ^{n+1}C_3(8)^3 + \dots$$

$$= 1 + (n + 1)(8) + 64 \cdot \left[^{n+1}C_2 + ^{n+1}C_3(8) + \dots\right]$$

$$= 1 + 8n + 8 + 64k = 8n + 9 + 64k$$

Subtracting $8n + 9$: $9^{n+1} - 8n - 9 = 64k$, which is clearly divisible by $64$!

7. Worked Examples — Easy to Hard

Example 1: Full Expansion of a Binomial Easy
Expand $\left(x^2 + \frac{3}{x}\right)^4$, $x \ne 0$.
Step 1: Write expansion with $n = 4$: $$\left(x^2 + \frac{3}{x}\right)^4 = ^4C_0 (x^2)^4 + ^4C_1 (x^2)^3\left(\frac{3}{x}\right) + ^4C_2 (x^2)^2\left(\frac{3}{x}\right)^2 + ^4C_3 (x^2)\left(\frac{3}{x}\right)^3 + ^4C_4 \left(\frac{3}{x}\right)^4$$
Step 2: Evaluate coefficients and powers:
  • $^4C_0 (x^2)^4 = 1 \cdot x^8 = x^8$
  • $^4C_1 (x^2)^3 \left(\frac{3}{x}\right) = 4 x^6 \left(\frac{3}{x}\right) = 12 x^5$
  • $^4C_2 (x^2)^2 \left(\frac{9}{x^2}\right) = 6 x^4 \left(\frac{9}{x^2}\right) = 54 x^2$
  • $^4C_3 (x^2) \left(\frac{27}{x^3}\right) = 4 x^2 \left(\frac{27}{x^3}\right) = \frac{108}{x}$
  • $^4C_4 \left(\frac{81}{x^4}\right) = \frac{81}{x^4}$

Final Expansion: $x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}$.

Example 2: Evaluating Numerical Power $(98)^5$ Easy
Using binomial theorem, evaluate $(98)^5$.
Step 1: Write $98$ as $(100 - 2)$: $$(98)^5 = (100 - 2)^5 = ^5C_0(100)^5 - ^5C_1(100)^4(2) + ^5C_2(100)^3(2^2) - ^5C_3(100)^2(2^3) + ^5C_4(100)(2^4) - ^5C_5(2^5)$$
Step 2: Calculate each term:
  • $= 10,000,000,000$
  • $- 5 \times 100,000,000 \times 2 = -1,000,000,000$
  • $+ 10 \times 1,000,000 \times 4 = +40,000,000$
  • $- 10 \times 10,000 \times 8 = -800,000$
  • $+ 5 \times 100 \times 16 = +8,000$
  • $- 32$
Step 3: Combine: $$10,040,008,000 - 1,000,800,032 = 9,039,207,968$$
Example 3: Finding a Specific Term ($13\text{th}$ Term) Medium
Find the $13\text{th}$ term in the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}$, $x \ne 0$.
Step 1: Formula for $T_{13}$ ($r = 12$): $$T_{13} = T_{12 + 1} = ^{18}C_{12} (9x)^{18 - 12} \left(-\frac{1}{3\sqrt{x}}\right)^{12}$$
Step 2: Simplify powers: Notice $(-1)^{12} = +1$. $$(9x)^6 = (3^2)^6 x^6 = 3^{12} x^6$$ $$\left(\frac{1}{3\sqrt{x}}\right)^{12} = \frac{1}{3^{12} (x^{1/2})^{12}} = \frac{1}{3^{12} x^6}$$
Step 3: Cancel factors: $$T_{13} = ^{18}C_{12} \cdot 3^{12} x^6 \cdot \frac{1}{3^{12} x^6} = ^{18}C_{12} = ^{18}C_6$$ $$= \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18,564$$
Example 4: Term Independent of $x$ Hard
Find the term independent of $x$ in the expansion of $\left(\frac{3x^2}{2} - \frac{1}{3x}\right)^9$.
Step 1: Write General Term $T_{r+1}$: $$T_{r+1} = ^9C_r \left(\frac{3x^2}{2}\right)^{9-r} \left(-\frac{1}{3x}\right)^r$$ $$= ^9C_r \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r (x^2)^{9-r} (x^{-1})^r$$ $$= ^9C_r \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{18 - 2r - r} = ^9C_r \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{18 - 3r}$$
Step 2: Set power of $x$ to $0$: $$18 - 3r = 0 \implies 3r = 18 \implies r = 6$$
Step 3: Calculate $T_7$: $$T_7 = ^9C_6 \left(\frac{3}{2}\right)^{9-6} \left(-\frac{1}{3}\right)^6 = ^9C_3 \left(\frac{3}{2}\right)^3 \left(\frac{1}{3^6}\right)$$ $$= 84 \times \frac{27}{8} \times \frac{1}{729} = 84 \times \frac{1}{8 \times 27} = \frac{84}{216} = \frac{7}{18}$$

Final Answer: $\frac{7}{18}$ (the 7th term).

Example 5: Finding Middle Term Medium
Find the middle term in the expansion of $\left(\frac{x}{3} + 9y\right)^{10}$.
Step 1: Identify middle term index: Here $n = 10$ is even. There is 1 middle term: $$\text{Middle term} = T_{\frac{10}{2} + 1} = T_6 \quad (\text{so } r = 5)$$
Step 2: Evaluate $T_6$: $$T_6 = ^{10}C_5 \left(\frac{x}{3}\right)^{10-5} (9y)^5 = ^{10}C_5 \frac{x^5}{3^5} \cdot (3^2)^5 y^5 = ^{10}C_5 \frac{x^5}{3^5} \cdot 3^{10} y^5$$ $$= ^{10}C_5 \cdot 3^5 x^5 y^5 = 252 \times 243 \times x^5 y^5 = 61,236 x^5 y^5$$

8. NCERT Key Exercise Questions Solved

NCERT Exercise 7.1 — Question 10

Q: Using Binomial Theorem, indicate which number is larger: $(1.1)^{10000}$ or $1000$.
View Full Solution

Write $1.1 = (1 + 0.1)$. Expand first two terms of $(1 + 0.1)^{10000}$:

$$(1 + 0.1)^{10000} = ^{10000}C_0 + ^{10000}C_1(0.1) + \text{other positive terms}$$

$$= 1 + 10000(0.1) + \text{positive terms}$$

$$= 1 + 1000 + \text{positive terms} = 1001 + \text{positive terms} > 1000$$

Conclusion: $(1.1)^{10000}$ is strictly larger than $1000$.

NCERT Exercise 7.1 — Question 11

Q: Find $(a + b)^4 - (a - b)^4$. Hence, evaluate $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$.
View Full Solution

$(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$.

$(a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$.

Subtracting: $(a + b)^4 - (a - b)^4 = 2[4a^3b + 4ab^3] = 8ab(a^2 + b^2)$.

Substitute $a = \sqrt{3}$ and $b = \sqrt{2}$:

$$8(\sqrt{3})(\sqrt{2})((\sqrt{3})^2 + (\sqrt{2})^2) = 8\sqrt{6}(3 + 2) = 8\sqrt{6}(5) = 40\sqrt{6}$$

NCERT Exercise 7.1 — Question 13

Q: Show that $9^{n+1} - 8n - 9$ is divisible by 64, whenever $n$ is a positive integer.
View Full Solution

Write $9 = 1 + 8$:

$$9^{n+1} = (1 + 8)^{n+1} = ^{n+1}C_0 + ^{n+1}C_1(8) + ^{n+1}C_2(8^2) + ^{n+1}C_3(8^3) + \dots + ^{n+1}C_{n+1}(8^{n+1})$$

$$= 1 + (n+1)8 + 64[^{n+1}C_2 + ^{n+1}C_3(8) + \dots]$$

$$= 1 + 8n + 8 + 64k = 8n + 9 + 64k$$

$$9^{n+1} - 8n - 9 = 64k$$

Since $k$ is an integer, $9^{n+1} - 8n - 9$ is divisible by 64. Hence proved!

9. Common Exam Traps & Pitfalls

Trap 1: The Off-By-One Error in $T_{r+1}$

To find the 5th term, DO NOT set $r = 5$! The 5th term is $T_{4+1}$, so $r = 4$. Setting $r = 5$ gives the 6th term and earns zero marks.

Trap 2: Middle Term Index When $n$ is Odd

When $n = 9$ (odd), students think the middle term is $9/2 = 4.5$. Remember: an expansion with $n=9$ has $9+1 = 10$ terms, so there are TWO middle terms: the 5th and 6th terms ($T_{(9+1)/2} = T_5$ and $T_{(9+3)/2} = T_6$).

Trap 3: Dropping the Negative Sign in Alternating Expansions

In $(x - 2y)^6$, the second term $b$ is $(-2y)$. When taking $b^r$, remember $(-2y)^r = (-1)^r 2^r y^r$. If $r$ is odd, the term is negative!

10. Quick Self-Check Quiz

Q1: How many terms are there in the expansion of $(2x - 3y)^{14}$?

Answer
$14 + 1 = 15$ terms.

Q2: What is the sum of coefficients in the expansion of $(1 - x)^{10}$?

Answer
Put $x = 1$: $(1 - 1)^{10} = 0^{10} = 0$.

Q3: Which term is the middle term in the expansion of $(x + a)^{12}$?

Answer
Since $n = 12$ is even, the middle term is $T_{12/2 + 1} = T_7$ (the 7th term).

Q4: What is the coefficient of $x^0$ (constant term) in $(x + 1/x)^2$?

Answer
$(x + 1/x)^2 = x^2 + 2(x)(1/x) + 1/x^2 = x^2 + 2 + 1/x^2$. The constant term is $2$.

Q5: What is the value of $^{15}C_0 + ^{15}C_1 + ^{15}C_2 + \dots + ^{15}C_{15}$?

Answer
$2^{15} = 32,768$.
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