MathematicsClass 11CBSE

How a Triangle of Numbers Expands Any Power of a Sum in Seconds

State the binomial theorem for positive integer powers and read its coefficients from Pascal's triangle, expand expressions such as (a + b)^n and (x + 1/x)^n, and use (1 + x)^n for approximations and divisibility proofs.

Why do powers of a sum follow a pattern?

You already know

(a + b)^{2} = a^{2} + 2 ab + b^{2} (a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

Multiplying out

(a + b)^{7}

by hand would take pages. The binomial theorem gives every term at once, because the coefficients

1, 2, 1

and

1, 3, 3, 1

are really combination numbers.

This chapter covers the theorem and Pascal's triangle, expanding binomials including

(x + \frac{1}{x})^{n}

, and using the special cases for approximations and divisibility.

What does the binomial theorem say, and how does Pascal's triangle give the coefficients?

**For a positive integer

n

(a + b)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r}

, which has

n + 1

terms whose coefficients form row

n

of Pascal's triangle.

Pascal's triangle** — each number is the sum of the two above it:

- Row

0

1

- Row

1

11

- Row

2

121

- Row

3

1331

- Row

4

14641

- Row

5

15101051

- Row

6

1615201561

The rule each number

=

sum of the two above is exactly

^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}

.

Worked example. The numbers in row

6

add to

1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 = 2^{6}

.

An everyday example. **Toss a coin

4

times**: the number of ways to get

0, 1, 2, 3, 4

heads is

1, 4, 6, 4, 1

— row

4

of the triangle.

The substance. The coefficients are symmetric, because

^{n} C_{r} =^{n} C_{n - r}

How do you expand (a + b)^n and expressions such as (x + 1/x)^n?

**Write

^{n} C_{r} a^{n - r} b^{r}

for

r = 0

n

, keeping signs and powers of each part, and use the general term

T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}

to find a single term.

Worked example 1.** Expand

(2 x - 3)^{4}

(2 x)^{4} + 4 (2 x)^{3} (- 3) + 6 (2 x)^{2} (- 3)^{2} + 4 (2 x) (- 3)^{3} + (- 3)^{4}

= 16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81

Worked example 2. Expand

(x + \frac{1}{x})^{6}

. The general term is

^{6} C_{r} x^{6 - r} x^{- r} =^{6} C_{r} x^{6 - 2 r}

x^{6} + 6 x^{4} + 15 x^{2} + 20 + \frac{15}{x ^{2}} + \frac{6}{x ^{4}} + \frac{1}{x ^{6}}

**The term independent of

x

20

**, where

6 - 2 r = 0

.

Worked example 3. The coefficient of

x^{5}

(x + 3)^{8}

comes from

r = 3

^{8} C_{3} \times 3^{3} = 56 \times 27 = 1512

An everyday example. **Arranging square tiles of sides

a

and

b

** into a big square of side

a + b

shows

(a + b)^{2}

as one

a^{2}

, two

ab

rectangles and one

b^{2}

.

The trap. **The

(r + 1)

th term uses

^{n} C_{r}

**, so the fourth term has

r = 3

How do you use (1 + x)^n and (1 - x)^n for approximations and divisibility results?

(1 + x)^{n} = 1 + n x +^{n} C_{2} x^{2} + \dots

and

(1 - x)^{n} = 1 - n x +^{n} C_{2} x^{2} - \dots

; when

x

is small, a few terms give an accurate approximation, and writing a number as

1 + k

reveals divisibility.

Worked example 1 — approximation.**

(0.99)^{5} = (1 - 0.01)^{5} \approx 1 - 0.05 + 10 (0.0001) = 0.951

Worked example 2 — exact large power.

(101)^{4} = (100 + 1)^{4} = 100000000 + 4000000 + 60000 + 400 + 1 = 104060401

Worked example 3 — comparison. Is

(1.1)^{10000}

larger than

1000

(1 + 0.1)^{10000} > 1 + 10000 \times 0.1 = 1001 > 1000

Worked example 4 — divisibility. Show that

6^{n} - 5 n - 1

is divisible by

25

6^{n} = (1 + 5)^{n} = 1 + 5 n +^{n} C_{2} 25 +^{n} C_{3} 125 + \dots

6^{n} - 5 n - 1 = 25 (^{n} C_{2} + 5^{n} C_{3} + \dots)

Check:

n = 3

gives

216 - 15 - 1 = 200 = 25 \times 8

.

An everyday example. **Monthly interest of

1%

for a year** grows a deposit by

(1.01)^{12} \approx 1 + 0.12 + 66 (0.0001) \approx 1.127

times.

The substance. **Dropping later terms is safe only when

x

is small**; for

(1.5)^{10}

the approximation fails badly.

Exam tip

What earns full marks on the binomial theorem?

**Write the general term first, then substitute

r

, keeping powers and signs of both parts of the binomial in brackets.

-

(a + b)^{n}

has

n + 1

terms
- General term**

T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}

- Negative second term: signs alternate
- Independent term: set the power of

x

to zero
- Approximations: keep enough terms for the accuracy asked
- Divisibility: write the base as

1 + k

and factor

k^{2}

from the later terms

The trap. Forgetting to raise the coefficient: **in

(2 x)^{3}

, the coefficient is

8

, not

2

.**

Did you know

Why do powers of 11 spell out rows of Pascal's triangle?

Look at the first few powers of

11

1 1^{1} = 111 1^{2} = 1211 1^{3} = 13311 1^{4} = 14641

**These digits are rows

1

4

of Pascal's triangle**, because

1 1^{n} = (10 + 1)^{n}

and each coefficient multiplies a power of

10

.

At

1 1^{5} = 161051

the pattern seems to break — but only because row

5

has two-digit numbers,

1, 5, 10, 10, 5, 1

, which carry into neighbouring places.

Exam relevance

How is the binomial theorem tested in JEE Main and JEE Advanced?

Binomial Theorem is a regular chapter for JEE Main and JEE Advanced, and binomial coefficients reappear in Class 12 Probability in the binomial distribution.

What gets asked. The general term, the middle term, the coefficient of a given power or the **term independent of

x

, sums of coefficients** found by substituting

x = 1

x = - 1

, and remainders of large powers, such as the remainder when a power of

7

is divided by a small number.

Question types. Numerical-value and multiple-choice questions, often needing careful index work.

The trap that costs marks. **Mixing up

T_{r}

and

T_{r + 1}

**, which shifts the term by one.

Key takeaways

What must you be able to do from this part?

- **

(a + b)^{n} = \sum^{n} C_{r} a^{n - r} b^{r}

**, with

n + 1

terms
- Pascal's triangle gives the coefficients; row

n

adds to

2^{n}

- **

(2 x - 3)^{4} = 16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81

(x + \frac{1}{x})^{6}

** has independent term

20

- General term

T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}

; coefficient of

x^{5}

(x + 3)^{8}

1512

- Approximations:

(0.99)^{5} \approx 0.951

; exact:

10 1^{4} = 104060401

- Divisibility:

6^{n} - 5 n - 1

is a multiple of

25

Use the binomial theorem to find

9 9^{3}

without a calculator, then check it by long multiplication.

Why do powers of a sum follow a pattern?

What does the binomial theorem say, and how does Pascal's triangle give the coefficients?

How do you expand (a + b)^n and expressions such as (x + 1/x)^n?

How do you use (1 + x)^n and (1 - x)^n for approximations and divisibility results?

What earns full marks on the binomial theorem?

Why do powers of 11 spell out rows of Pascal's triangle?

How is the binomial theorem tested in JEE Main and JEE Advanced?

What must you be able to do from this part?

Related study notes

Ready to put this into practice?