MathsCBSE Class 11 · Chapter 7

Binomial Theorem Class 11: Notes, Pascal's Triangle and Solved Questions

Updated 14 October 2026 · Fact-checked

Binomial Theorem is Chapter 7 of CBSE Class 11 Maths. For 2026-27 it covers the history of the theorem, Pascal's triangle, the statement and proof of the binomial theorem for positive integral indices, special cases such as $(1 + x)^n$ and $(x - y)^n$, and simple applications to large powers, comparisons and divisibility.

What Binomial Theorem is about

You already know $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ and $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . But expanding $(a + b)^{7}$ by repeated multiplication is slow and error-prone. The binomial theorem gives every term of $(a + b)^{n}$ directly.

The coefficients form Pascal's triangle: each row starts and ends with 1, and every other entry is the sum of the two entries above it. These numbers are exactly the combinations $^{n} C_{r}$ from the previous chapter, which gives the theorem
$(a + b)^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + \dots +^{n} C_{n} b^{n}$

The expansion has $n + 1$ terms; the power of a falls from n to 0 while the power of b rises from 0 to n. Putting special values gives useful results such as $^{n} C_{0} +^{n} C_{1} + \dots +^{n} C_{n} = 2^{n}$ . The theorem lets you compute powers like $9 9^{4}$ quickly, compare numbers like $(1.02)^{500}$ and 10, approximate small powers, and prove divisibility results such as ' $7^{n} - 6 n - 1$ is divisible by 36'. For 2026-27, the general and middle terms are assessed only formatively and are not in the annual exam.

Weightage: This chapter is part of Unit II, Algebra, which the syllabus gives 25 of the 80 theory marks in the school annual exam (CBSE adds that there is no fixed chapter-wise weightage).Source: CBSE Mathematics (041) syllabus 2026-27, Class XI course structure, Unit II Algebra

Key concepts

Pascal's triangle

NCERT section 7.2

Write the coefficients of $(a + b)^{n}$ for n = 0, 1, 2, … in rows:

Index n	Coefficients
0	1
1	1, 1
2	1, 2, 1
3	1, 3, 3, 1
4	1, 4, 6, 4, 1
5	1, 5, 10, 10, 5, 1
6	1, 6, 15, 20, 15, 6, 1

Rules: each row begins and ends with 1; every inner number is the sum of the two numbers just above it (for example 15 = 5 + 10). Each row is symmetric.

This array was known long before Pascal: Pingala's *Meru Prastara* in ancient India described the same pattern.

The entries are binomial coefficients: the r-th entry (counting from 0) in row n is $^{n} C_{r}$ . The 'add the two above' rule is exactly Pascal's rule $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ .

Worked example

Use Pascal's triangle to expand $(x + 3)^{4}$ . [2 marks]

Try it first, then show the solutionSolution

Row 4: 1, 4, 6, 4, 1
$(x + 3)^{4} = x^{4} + 4 x^{3} (3) + 6 x^{2} (9) + 4 x (27) + 81 = x^{4} + 12 x^{3} + 54 x^{2} + 108 x + 81$

Answer: $x^{4} + 12 x^{3} + 54 x^{2} + 108 x + 81$

Statement of the binomial theorem

NCERT section 7.2.1

For any positive integer n and any numbers a, b:
$(a + b)^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b +^{n} C_{2} a^{n - 2} b^{2} + \dots +^{n} C_{n - 1} a b^{n - 1} +^{n} C_{n} b^{n} = \sum_{k = 0}^{n}^{n} C_{k} a^{n - k} b^{k}$

Observations:
1. There are $n + 1$ terms, one more than the index.
2. The power of a decreases by 1 from n to 0; the power of b increases by 1 from 0 to n.
3. In every term the two powers add up to n.
4. The coefficients $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ are the binomial coefficients; equidistant coefficients from the two ends are equal, since $^{n} C_{r} =^{n} C_{n - r}$ .

**Why $^{n} C_{k}$ ?** Multiplying out $(a + b) (a + b) \dots (a + b)$ (n brackets), a term $a^{n - k} b^{k}$ comes from choosing b from k of the n brackets and a from the rest. There are $^{n} C_{k}$ ways to choose those k brackets.

Worked example

Expand $(2 x - y)^{5}$ . [3 marks]

Try it first, then show the solutionSolution

Coefficients: 1, 5, 10, 10, 5, 1, with a = 2x and b = −y
$(2 x)^{5} + 5 (2 x)^{4} (- y) + 10 (2 x)^{3} y^{2} + 10 (2 x)^{2} (- y)^{3} + 5 (2 x) y^{4} + (- y)^{5}$
$= 32 x^{5} - 80 x^{4} y + 80 x^{3} y^{2} - 40 x^{2} y^{3} + 10 x y^{4} - y^{5}$

Answer: $32 x^{5} - 80 x^{4} y + 80 x^{3} y^{2} - 40 x^{2} y^{3} + 10 x y^{4} - y^{5}$

Proof for positive integral index

NCERT section 7.2.1

NCERT proves the theorem by mathematical induction (proving it for n = 1, then showing that truth for n = k forces truth for n = k + 1).

Step 1 (n = 1): $(a + b)^{1} =^{1} C_{0} a +^{1} C_{1} b = a + b$ . True.

Step 2: assume $(a + b)^{k} = \sum_{r = 0}^{k}^{k} C_{r} a^{k - r} b^{r}$ for some positive integer k.

Step 3: $(a + b)^{k + 1} = (a + b) (a + b)^{k}$ . Multiply each term of the assumed expansion by a and by b, then collect the terms with the same powers. The term $a^{k + 1 - r} b^{r}$ gets the coefficient $^{k} C_{r} +^{k} C_{r - 1}$ , which equals $^{k + 1} C_{r}$ by Pascal's rule. The first and last coefficients are $^{k} C_{0} = 1 =^{k + 1} C_{0}$ and $^{k} C_{k} = 1 =^{k + 1} C_{k + 1}$ .

So the formula holds for $k + 1$ . By induction it holds for every positive integer n.

(The induction chapter itself is formative-only in 2026-27, but this proof is part of the syllabus statement 'statement and proof of the binomial theorem'.)

Worked example

In the proof, which identity is used to combine $^{k} C_{r}$ and $^{k} C_{r - 1}$ ? Verify it for k = 5, r = 2. [1 mark]

Try it first, then show the solutionSolution

Pascal's rule: $^{k} C_{r} +^{k} C_{r - 1} =^{k + 1} C_{r}$
$^{5} C_{2} +^{5} C_{1} = 10 + 5 = 15 =^{6} C_{2}$ ✓

Answer: Pascal's rule; 15 = 15.

Special cases and sums of coefficients

NCERT section 7.2.2

1. b replaced by −y: $(x - y)^{n} =^{n} C_{0} x^{n} -^{n} C_{1} x^{n - 1} y +^{n} C_{2} x^{n - 2} y^{2} - \dots + (- 1)^{n}^{n} C_{n} y^{n}$ . The signs alternate.
2. a = 1: $(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + \dots +^{n} C_{n} x^{n}$ .
3. Put x = 1: $^{n} C_{0} +^{n} C_{1} + \dots +^{n} C_{n} = 2^{n}$ (the sum of each row of Pascal's triangle).
4. ** $(1 - x)^{n}$ with x = 1:** $^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - \dots + (- 1)^{n}^{n} C_{n} = 0$ .
5. Adding and subtracting: in $(a + b)^{n} + (a - b)^{n}$ the terms with odd powers of b cancel and the others double; in $(a + b)^{n} - (a - b)^{n}$ the terms with even powers of b cancel.

Sum of the coefficients of any expansion in x and y: put x = y = 1.

Worked example

Find $(a + b)^{5} - (a - b)^{5}$ and hence evaluate $(2 + 1)^{5} - (2 - 1)^{5}$ . [3 marks]

Try it first, then show the solutionSolution

Terms with even powers of b cancel; odd ones double: $2 (5 a^{4} b + 10 a^{2} b^{3} + b^{5}) = 10 a^{4} b + 20 a^{2} b^{3} + 2 b^{5}$
Put $a = 2$ , b = 1: $a^{4} = 4$ , $a^{2} = 2$ , so $10 (4) + 20 (2) + 2$
$= 40 + 40 + 2 = 82$

Answer: $10 a^{4} b + 20 a^{2} b^{3} + 2 b^{5}$ ; 82

Computing powers and comparing numbers

NCERT section 7.2 (Examples 2 and 3)

Large powers: write the base as a round number plus or minus a small one, then expand. For $9 9^{4}$ , use $(100 - 1)^{4}$ .

Comparisons: for $(1 + x)^{n}$ with $x > 0$ , all terms are positive, so
$(1 + x)^{n} > 1 + n x$ (for $n \geq 2$ )
This quickly shows that a power is bigger than some number.

Approximations: if x is small, the higher powers of x are tiny, so a few terms give a good estimate: $(1 + x)^{n} \approx 1 + n x + \frac{n ( n - 1 )}{2} x^{2}$ .

Worked example

Using the binomial theorem, evaluate $9 9^{4}$ . [2 marks]

Try it first, then show the solutionSolution

$(100 - 1)^{4} = 10 0^{4} - 4 (100)^{3} + 6 (100)^{2} - 4 (100) + 1$
$= 100000000 - 4000000 + 60000 - 400 + 1 = 96059601$

Answer: 96059601

Divisibility and remainders

NCERT section 7.2 (Example 4)

To show that an expression like $c^{n}$ minus something is divisible by $m$ , write c as $1 + d$ (or $m \pm 1$ ) and expand:
$(1 + d)^{n} = 1 + n d +^{n} C_{2} d^{2} +^{n} C_{3} d^{3} + \dots$
Every term from $d^{2}$ onwards contains the factor $d^{2}$ .

Example pattern: $7^{n} = (1 + 6)^{n} = 1 + 6 n + 36 (^{n} C_{2} + 6^{n} C_{3} + \dots)$ . So $7^{n} - 6 n - 1 = 36 k$ for a whole number k: divisible by 36.

Remainders: $2^{30} = 8^{10} = (7 + 1)^{10}$ ; every term except the last 1 contains 7, so the remainder on division by 7 is 1.

Factor result: $a^{n} - b^{n}$ is divisible by $a - b$ : write $a = (a - b) + b$ and expand $a^{n}$ ; every term except $b^{n}$ contains $(a - b)$ .

Worked example

Prove that $7^{n} - 6 n - 1$ is divisible by 36 for every positive integer n. [3 marks]

Try it first, then show the solutionSolution

$7^{n} = (1 + 6)^{n} = 1 +^{n} C_{1} (6) +^{n} C_{2} (6^{2}) +^{n} C_{3} (6^{3}) + \dots + 6^{n}$
$7^{n} - 6 n - 1 = 6^{2} [^{n} C_{2} +^{n} C_{3} (6) + \dots + 6^{n - 2}]$ (for n = 1 the expression is 0, which is also divisible)
The bracket is a whole number k, so the expression equals 36k

Answer: Proved: divisible by 36.

Maths in real life: Binomial Theorem

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Compound growth approximation

Suppose ₹10000 is invested at 2% per year compound interest for 5 years, and you have no calculator.

Estimate the amount using the first three terms of $(1.02)^{5}$ .

Try it first, then show the solutionSolution

$(1 + 0.02)^{5} \approx 1 + 5 (0.02) + 10 (0.02)^{2} = 1 + 0.1 + 0.004 = 1.104$
Amount ≈ $10000 \times 1.104 = 11040$ (the exact value is about ₹11040.81)

Answer: About ₹11040

Memory hook: Small rate: the first three terms are nearly everything.

Counting with binomial coefficients

Imagine tossing a fair coin 5 times and writing each outcome as a string of H and T.

How many outcomes have exactly 2 heads? Relate this to the expansion of $(H + T)^{5}$ .

Try it first, then show the solutionSolution

Choose which 2 of the 5 tosses are heads: $^{5} C_{2} = 10$
In $(H + T)^{5}$ the term $H^{2} T^{3}$ has coefficient $^{5} C_{2} = 10$ , and all coefficients add up to $2^{5} = 32$ outcomes

Answer: 10 of the 32 outcomes

Memory hook: The coefficient counts the ways to pick which brackets give b.

Comparing growth

Suppose a town's population grows by 2% every year for 500 years (a thought experiment).

Show that it becomes more than 10 times as large.

Try it first, then show the solutionSolution

Growth factor $(1.02)^{500} = 1 + 500 (0.02) + positive terms$
$= 1 + 10 + \dots > 11 > 10$

Answer: More than 10 times (in fact far more)

Memory hook: $(1 + x)^{n} > 1 + n x$ for x > 0.

Topics in this chapter

Binomial Theorem formula sheet

Binomial theorem: $(a + b)^{n} = \sum_{k = 0}^{n}^{n} C_{k} a^{n - k} b^{k}$
With a minus sign: $(x - y)^{n} = \sum_{k = 0}^{n} (- 1)^{k}^{n} C_{k} x^{n - k} y^{k}$
Powers of (1 + x): $(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + \dots +^{n} C_{n} x^{n}$
Sums of binomial coefficients: $\sum_{k = 0}^{n}^{n} C_{k} = 2^{n}, \sum_{k = 0}^{n} (- 1)^{k}^{n} C_{k} = 0$
Pascal's rule: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
Approximation for small x: $(1 + x)^{n} \approx 1 + n x + \frac{n ( n - 1 )}{2} x^{2}$; For x > 0 and n ≥ 2, $(1 + x)^{n} > 1 + n x$ .

How the school exam asks Binomial Theorem

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Expand a binomial	Expand expressions like $(1 - 2 x)^{5}$ , $(\frac{x}{2} - \frac{2}{x})^{4}$ or $(x + \frac{1}{x})^{6}$ (Exercise 7.1, Q1-5).	Write the coefficient row, keep signs inside the second term, simplify powers carefully.
Evaluate a large power	Find $9 6^{3}$ , $10 2^{5}$ , $9 9^{5}$ using the theorem.	Split into 100 ± small number; most terms are easy powers of 10.
Compare or approximate	Which is larger, $(1.1)^{10000}$ or 1000? Approximate $(0.99)^{5}$ to three terms.	Keep the first two or three terms; for comparison, the rest are positive.
Sum and difference of expansions	Find $(a + b)^{n} \pm (a - b)^{n}$ and evaluate expressions with surds.	Write only the surviving terms, doubled.
Divisibility and proofs	Show $9^{n + 1} - 8 n - 9$ is divisible by 64; prove $\sum 3^{r}^{n} C_{r} = 4^{n}$ ; show $a - b$ divides $a^{n} - b^{n}$ .	Write the base as 1 + d or as (a − b) + b, expand, factor out.

Tricks to remember Binomial Theorem

Short memory aids. Say each one aloud once, then use it on a question the same day.

Down goes a, up goes b
Start with $a^{n}$ ; each step lower a's power by 1 and raise b's by 1; powers always add to n.
Use it when: Writing terms in order.
n + 1 terms
One more term than the power: $(a + b)^{6}$ has 7 terms.
Use it when: Counting terms.
Hockey-stick adding
Each Pascal entry = left-above + right-above.
Use it when: Building rows quickly.
Put 1 to sum
Sum of coefficients: put every variable equal to 1.
Use it when: Sum-of-coefficient questions.
Keep the minus inside
For $(a - b)^{n}$ , treat the second term as $(- b)$ : odd powers negative, even powers positive.
Use it when: Signs in expansions.

Binomial Theorem: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Number of terms · 1 mark · MCQExam level

The number of terms in the expansion of $(2 x + 3 y)^{8}$ is:
(a) 8
(b) 9
(c) 10
(d) 16

Try it first, then show the solutionSolution

Index 8 gives 8 + 1 terms.

Answer: (b) 9

Example 2: Sum of coefficients · 1 mark · Assertion-reasonExam level

Assertion (A): The sum of the binomial coefficients in $(1 + x)^{7}$ is 128.
Reason (R): Putting x = 1 in $(1 + x)^{n}$ gives $\sum^{n} C_{r} = 2^{n}$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

$2^{7} = 128$ , which follows from R.

Answer: (a)

Example 3: Expansion with reciprocals · 3 marks · Short answerHarder exam level

Expand $(x + \frac{1}{x})^{4}$ and hence find $(x + \frac{1}{x})^{4} + (x - \frac{1}{x})^{4}$ . [3 marks]

Try it first, then show the solutionSolution

$(x + \frac{1}{x})^{4} = x^{4} + 4 x^{2} + 6 + \frac{4}{x ^{2}} + \frac{1}{x ^{4}}$
$(x - \frac{1}{x})^{4} = x^{4} - 4 x^{2} + 6 - \frac{4}{x ^{2}} + \frac{1}{x ^{4}}$
Sum $= 2 x^{4} + 12 + \frac{2}{x ^{4}}$

Answer: $2 x^{4} + 12 + \frac{2}{x ^{4}}$

Example 4: A large power · 2 marks · Very short answerHarder exam level

Using the binomial theorem, evaluate $10 3^{3}$ . [2 marks]

Try it first, then show the solutionSolution

$(100 + 3)^{3} = 10 0^{3} + 3 (10 0^{2}) (3) + 3 (100) (9) + 27$
$= 1000000 + 90000 + 2700 + 27 = 1092727$

Answer: 1092727

Example 5: Approximation · 2 marks · Very short answerHarder exam level

Find an approximation of $(0.98)^{6}$ using the first three terms of its expansion, and say whether it is an over- or under-estimate. [2 marks]

Try it first, then show the solutionSolution

$(1 - 0.02)^{6} \approx 1 - 6 (0.02) + 15 (0.0004) = 1 - 0.12 + 0.006 = 0.886$
The next term, $- 20 (0.02)^{3} = - 0.00016$ , is negative and the later ones are far smaller, so 0.886 is a slight over-estimate (true value 0.88584…)

Answer: 0.886, slightly too large

Example 6: Remainder · 2 marks · Very short answerHarder exam level

Find the remainder when $2^{30}$ is divided by 7. [2 marks]

Try it first, then show the solutionSolution

$2^{30} = (2^{3})^{10} = 8^{10} = (7 + 1)^{10}$
Every term of the expansion except the last ( $1^{10} = 1$ ) contains a factor 7, so the remainder is 1

Answer: 1

Example 7: Which is larger? · 3 marks · Short answerHOTS / competency

Without calculating, show that $10 1^{50} > 9 9^{50} + 10 0^{50}$ . [3 marks]

Try it first, then show the solutionSolution

$10 1^{50} - 9 9^{50} = (100 + 1)^{50} - (100 - 1)^{50}$ : the even-power terms cancel and the odd ones double
$= 2 [^{50} C_{1} 10 0^{49} +^{50} C_{3} 10 0^{47} + \dots] > 2 \times 50 \times 10 0^{49} = 10 0^{50}$
So $10 1^{50} - 9 9^{50} > 10 0^{50}$ , i.e. $10 1^{50} > 9 9^{50} + 10 0^{50}$

Answer: Proved.

Example 8: Divisibility by 49 · 3 marks · Short answerHOTS / competency

Prove that $2^{3 n} - 7 n - 1$ is divisible by 49 for every positive integer n. [3 marks]

Try it first, then show the solutionSolution

$2^{3 n} = 8^{n} = (1 + 7)^{n} = 1 +^{n} C_{1} (7) +^{n} C_{2} (7^{2}) + \dots + 7^{n}$
$2^{3 n} - 7 n - 1 = 7^{2} [^{n} C_{2} +^{n} C_{3} (7) + \dots + 7^{n - 2}]$ (for n = 1 the expression is 0)
The bracket is a whole number k, so the expression equals 49k

Answer: Proved.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Pascal's triangle in a quiz

Suppose a quiz club builds Pascal's triangle with tiles, one row at a time, starting from row 0 (a single 1).

(i) · 1 mark
Write row 6.
Answer
Answer: 1, 6, 15, 20, 15, 6, 1
Add pairs from row 5: 1, 5, 10, 10, 5, 1.
(ii) · 1 mark
What is the sum of the numbers in row 6?
Answer
Answer: 64
$2^{6}$
(iii) · 1 mark
Use row 6 to find the coefficient of $x^{3}$ in $(1 + x)^{6}$ .
Answer
Answer: 20
$^{6} C_{3}$
(iv) · 1 mark
What is $1 - 6 + 15 - 20 + 15 - 6 + 1$ ?
Answer
Answer: 0
$(1 - 1)^{6} = 0$

Savings estimate

Suppose Meera's savings of ₹20000 grow at 1% per month, compounded monthly, for 3 months.

(i) · 1 mark
Write the amount as a power.
Answer
Answer: $20000 (1.01)^{3}$
Multiply by 1.01 each month.
(ii) · 1 mark
Expand $(1.01)^{3}$ fully.
Answer
Answer: 1.030301
$1 + 3 (0.01) + 3 (0.0001) + 0.000001$
(iii) · 1 mark
Find the amount.
Answer
Answer: ₹20606.02
$20000 \times 1.030301$
(iv) · 1 mark
How much error does using only the first two terms cause?
Answer
Answer: About ₹6.02
Two terms give 1.03 and ₹20600; the difference is $20000 \times 0.000301$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Expand $(1 - 3 x)^{4}$ . [2 marks] (2 marks)

Step	Marks
Coefficients 1, 4, 6, 4, 1 with b = −3x: $1 + 4 (- 3 x) + 6 (9 x^{2}) + 4 (- 27 x^{3}) + 81 x^{4}$	1
$= 1 - 12 x + 54 x^{2} - 108 x^{3} + 81 x^{4}$	1

Find $(x + 1)^{5} + (x - 1)^{5}$ and hence evaluate $(3 + 1)^{5} + (3 - 1)^{5}$ . [3 marks] (3 marks)

Step	Marks
The terms with odd powers of the second term (±1) cancel and the rest double: $2 (x^{5} + 10 x^{3} + 5 x)$	1
Put $x = 3$ : $x^{5} = 93$ , $x^{3} = 33$	1
$2 (93 + 303 + 53) = 2 \times 443 = 883$	1

(a) State the binomial theorem for a positive integer n. (b) Expand $(\frac{x}{3} - \frac{3}{x})^{4}$ . (c) Use the theorem with $(1 + x)^{n}$ to show that $\sum_{r = 0}^{n} 2^{r}^{n} C_{r} = 3^{n}$ . [5 marks] (5 marks)

Step	Marks
(a) $(a + b)^{n} = \sum_{k = 0}^{n}^{n} C_{k} a^{n - k} b^{k}$ for every positive integer n	1
(b) Terms: $\frac{x ^{4}}{81}$ , $4 \cdot \frac{x ^{3}}{27} \cdot (- \frac{3}{x}) = - \frac{4 x ^{2}}{9}$ , $6 \cdot \frac{x ^{2}}{9} \cdot \frac{9}{x ^{2}} = 6$	1
$4 \cdot \frac{x}{3} \cdot (- \frac{27}{x ^{3}}) = - \frac{36}{x ^{2}}$ , $\frac{81}{x ^{4}}$ ; so the expansion is $\frac{x ^{4}}{81} - \frac{4 x ^{2}}{9} + 6 - \frac{36}{x ^{2}} + \frac{81}{x ^{4}}$	1
(c) $(1 + x)^{n} = \sum_{r = 0}^{n}^{n} C_{r} x^{r}$	1
Put x = 2: $\sum_{r = 0}^{n}^{n} C_{r} 2^{r} = (1 + 2)^{n} = 3^{n}$	1

Where students lose marks in Binomial Theorem

The slips examiners see most often, and what to write instead.

Mistake: Forgetting to raise the coefficient inside a term
Applying the power only to the variable.
Wrong working
$(2 x)^{3} = 2 x^{3}$
Right working
$(2 x)^{3} = 8 x^{3}$
Do this: Put brackets round each whole term before raising it to a power.
Usually costs: 1-2 marks.
Mistake: Losing the alternating signs
Writing all plus signs for $(a - b)^{n}$ .
Do this: Odd powers of (−b) are negative.
Usually costs: 1 mark.
Mistake: Wrong row of Pascal's triangle
Starting the count at row 1.
Do this: The row for index n has n + 1 numbers; the row for 5 is 1, 5, 10, 10, 5, 1.
Usually costs: Whole expansion.
Mistake: Arithmetic slips in large powers
Too many zeros.
Do this: Write each term with its powers of 10 in a column before adding.
Usually costs: 1 mark.
Mistake: Writing the general-term formula as the method
Old guides.
Do this: General and middle terms are formative-only for 2026-27; in the annual exam, expand and read off terms.
Usually costs: Usually none, but the expected method is expansion.
Mistake: Simplifying $x^{a} \cdot \frac{1}{x ^{b}}$ wrongly
Adding instead of subtracting powers.
Do this: $x^{a} \cdot x^{- b} = x^{a - b}$ ; check each term's power of x.
Usually costs: 1 mark.

NCERT exercise guide: Binomial Theorem

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 7.1

What it tests: Expanding binomials with numbers, negative terms and reciprocals; evaluating large powers; comparisons; sums and differences of expansions with surds; divisibility; sums of the form $\sum k^{r}^{n} C_{r}$ .

How to approach it: Pascal's row or $^{n} C_{r}$ ; split numbers as 100 ± k; keep only surviving terms; write the base as 1 + d.

Same kind as Exercise 7.1

Using the binomial theorem, show which is larger: $(1.02)^{500}$ or 10.

Show the solutionSolution

$(1 + 0.02)^{500} = 1 + 500 (0.02) + other positive terms$
$= 1 + 10 + \dots = 11 + \dots > 10$

Answer: $(1.02)^{500}$ is larger.

Miscellaneous Exercise on Chapter 7

What it tests: $a - b$ divides $a^{n} - b^{n}$ ; surd evaluations; approximations with three terms; expansions of expressions with two or three terms.

How to approach it: Write a = (a − b) + b; group a trinomial as a binomial of two parts.

Same kind as Miscellaneous Exercise

Find an approximation of $(1.02)^{6}$ using the first three terms of its expansion.

Show the solutionSolution

$(1 + 0.02)^{6} \approx 1 + 6 (0.02) + 15 (0.02)^{2}$
$= 1 + 0.12 + 0.006 = 1.126$

Answer: 1.126 (exact value 1.12616…)

Practice questions on Binomial Theorem

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The coefficient of $x^{3}$ in $(1 + 2 x)^{5}$ is:
(a) 10
(b) 40
(c) 80
(d) 32
Hint
$^{5} C_{3} \times 2^{3}$ .
Answer
(c) 80
Q2 · 1 mark · MCQExam level
$^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - \dots + (- 1)^{n}^{n} C_{n}$ equals:
(a) $2^{n}$
(b) 0
(c) 1
(d) $2^{n - 1}$
Hint
$(1 - 1)^{n}$ .
Answer
(b) 0
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): In $(a + b)^{n}$ the coefficients equidistant from the ends are equal.
Reason (R): $^{n} C_{r} =^{n} C_{n - r}$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
The r-th term from each end.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Expand $(x + 2)^{5}$ using Pascal's triangle.
Hint
Row 5.
Answer
$x^{5} + 10 x^{4} + 40 x^{3} + 80 x^{2} + 80 x + 32$ .
Q5 · 2 marks · Very short answerHarder exam level
Evaluate $9 8^{4}$ using the binomial theorem.
Hint
$(100 - 2)^{4}$ .
Answer
$100000000 - 4 (1000000) (2) + 6 (10000) (4) - 4 (100) (8) + 16 = 100000000 - 8000000 + 240000 - 3200 + 16 = 92236816$ .
Q6 · 2 marks · Very short answerHarder exam level
How many terms are in the expansion of $(1 + x)^{10} + (1 - x)^{10}$ after simplification?
Hint
Odd powers cancel.
Answer
Only even powers $x^{0}, x^{2}, \dots, x^{10}$ survive: 6 terms.
Q7 · 1 mark · Very short answerHarder exam level
Find the sum of the coefficients in the expansion of $(3 x - 2 y)^{7}$ .
Hint
x = y = 1.
Answer
$(3 - 2)^{7} = 1$ .
Q8 · 3 marks · Short answerHarder exam level
Show that $1 1^{n} - 10 n - 1$ is divisible by 100 for every positive integer n.
Hint
$11 = 1 + 10$ .
Answer
$1 1^{n} = 1 + 10 n + 1 0^{2} (^{n} C_{2} + 10^{n} C_{3} + \dots)$ , so $1 1^{n} - 10 n - 1 = 100 k$ with k a whole number.
Q9 · 3 marks · Short answerHarder exam level
Expand $(x^{2} + x - 1)^{3}$ by treating it as $[(x^{2} + x) - 1]^{3}$ .
Hint
Let $u = x^{2} + x$ .
Answer
$u^{3} - 3 u^{2} + 3 u - 1$ with $u^{3} = x^{6} + 3 x^{5} + 3 x^{4} + x^{3}$ , $u^{2} = x^{4} + 2 x^{3} + x^{2}$ : $x^{6} + 3 x^{5} + 3 x^{4} + x^{3} - 3 x^{4} - 6 x^{3} - 3 x^{2} + 3 x^{2} + 3 x - 1 = x^{6} + 3 x^{5} - 5 x^{3} + 3 x - 1$ .
Q10 · 3 marks · Short answerHOTS / competency
Using the binomial theorem, show that $(1.1)^{100} > 11$ without a calculator. Then decide, with reasons, whether $(1.1)^{100} > 100$ as well.
Hint
Include the third term.
Answer
$(1 + 0.1)^{100} > 1 + 100 (0.1) = 11$ . Adding the third term: $^{100} C_{2} (0.01) = 4950 \times 0.01 = 49.5$ , so it exceeds 60.5; the fourth term $^{100} C_{3} (0.001) = 161700 \times 0.001 = 161.7$ , so the total exceeds 222 > 100. Yes.
Q11 · 2 marks · Very short answerHOTS / competency
Explain, using the idea of choosing brackets, why the coefficient of $a^{2} b^{3}$ in $(a + b)^{5}$ is 10.
Hint
Which brackets give b?
Answer
Each term of the product of five brackets picks a or b from each bracket. To get $a^{2} b^{3}$ , pick b from 3 of the 5 brackets (and a from the rest): $^{5} C_{3} = 10$ ways, each giving one copy of $a^{2} b^{3}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.How many terms are in $(a + b)^{9}$ ?Show answer
10.
2.Write row 5 of Pascal's triangle.Show answer
1, 5, 10, 10, 5, 1.
3.What is the sum $^{n} C_{0} +^{n} C_{1} + \dots +^{n} C_{n}$ ?Show answer
$2^{n}$ .
4.In $(a + b)^{n}$ , what do the powers of a and b add to in each term?Show answer
n.
5.Sum of the coefficients of $(2 x - 3 y)^{5}$ ?Show answer
$(2 - 3)^{5} = - 1$ .
6.Which identity drives the induction proof?Show answer
$^{k} C_{r} +^{k} C_{r - 1} =^{k + 1} C_{r}$ .
7.Approximate $(1.02)^{6}$ to three terms.Show answer
$1 + 0.12 + 0.006 = 1.126$ .
8.What is the remainder when $2^{30}$ is divided by 7?Show answer
1.

Best order to study it

Revise $^{n} C_{r}$ and Pascal's rule from Chapter 6.
Build Pascal's triangle up to row 7 and expand small powers with it.
Learn the statement of the theorem and its observations.
Understand the proof by induction and the 'choose the brackets' explanation.
Practise expansions with negative and fractional terms.
Learn the special cases and sums of coefficients.
Practise large powers, comparisons and approximations.
Finish with divisibility and remainder proofs.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write any row of Pascal's triangle up to row 7?
Can you state the binomial theorem in sigma form?
Can you prove it by induction using Pascal's rule?
Can you expand $(x - \frac{1}{x})^{5}$ correctly with signs?
Can you find $(a + b)^{n} + (a - b)^{n}$ quickly?
Can you evaluate $10 2^{4}$ or $9 8^{4}$ using the theorem?
Can you show $(1.1)^{10000} > 1000$ ?
Can you approximate $(0.99)^{5}$ using three terms?
Can you prove a divisibility result like $7^{n} - 6 n - 1$ by 36?
Can you show that $\sum 3^{r}^{n} C_{r} = 4^{n}$ ?

Binomial Theorem: frequently asked questions

Are general and middle terms in the 2026-27 exam?

No. The NCERT section on general and middle terms was removed, and CBSE lists them as assessed only formatively. The annual exam covers the statement, proof, Pascal's triangle and simple applications.

Do I need to know mathematical induction for the proof?

The NCERT proof uses induction, so you should follow its three steps. The induction chapter itself is formative-only, so you will not be asked general induction proofs, but the binomial theorem's proof is part of the syllabus.

Does the theorem work for negative or fractional powers?

There is a version for any real index, but it gives an infinite series and is beyond Class 11. This chapter is only about positive integer powers.

Why is it called 'binomial'?

A binomial is an expression with two terms, like a + b. The theorem expands any positive integer power of such a two-term expression.

How is the Class 11 Maths exam set?

It is a school annual exam of 80 marks (3 hours) plus 20 marks of internal assessment; Unit II, Algebra, which includes this chapter, carries 25 marks. CBSE gives no Class 11 sample paper, so schools usually mix MCQs, assertion-reason, short and long answers and case-based questions.

Based on: CBSE Mathematics (041) syllabus 2026-27 (Classes XI-XII); NCERT Mathematics, Textbook for Class XI, Chapter 7 (Reprint 2026-27)