Introduction to Binomial Expansion

MathsBinomial Theorem And Mathematical InductionFor JEE aspirants

The Binomial Theorem gives a direct formula to expand any power of a binomial expression $(a + x)^{n}$ for a positive integer $n$ , without multiplying it out by hand. For JEE and NEET aspirants (JEE Main, JEE Advanced), it is a Class 11 chapter whose ideas power problems on coefficients, middle terms, greatest terms, divisibility, and approximations. The expansion is $(a + x)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} x^{r}$ , has exactly $n + 1$ terms, and its general term is $T_{r + 1} =^{n} C_{r} a^{n - r} x^{r}$ . Mastering this expansion is the foundation for every follow-on binomial topic.

Key Formulas - Quick Reference

Positive-index expansion: $(a + x)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} x^{r}$
General term: $T_{r + 1} =^{n} C_{r} a^{n - r} x^{r}$
Alternating-sign form: $(a - x)^{n} = \sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} a^{n - r} x^{r}$
Middle term ( $n$ even): $T_{n /2 + 1} =^{n} C_{n /2} a^{n /2} x^{n /2}$
Middle terms ( $n$ odd): $T_{(n + 1) /2}$ and $T_{(n + 3) /2}$
Greatest binomial coefficient: $^{n} C_{n /2}$ (even $n$ ); $^{n} C_{(n \pm 1) /2}$ (odd $n$ )
Numerically greatest term index: $m = \frac{( n + 1 ) ∣ x ∣}{1 + ∣ x ∣}$ in $(1 + x)^{n}$
$p$ -th term from end = $(n - p + 2)$ -th term from beginning

1. What is a Binomial Expression?

An algebraic expression containing exactly two terms is called a binomial expression. Examples include $(a + b)$ , $(2 x - 3 y)$ , $(x + \frac{1}{y})$ , $(x + \frac{3}{x})$ , and $(\frac{2}{x} - \frac{1}{x ^{2}})$ .

2. Binomial Theorem for a Positive Integer Index

The formula that lets us expand any power of a binomial expression as a finite series is the Binomial Theorem. For a positive integer $n$ :

(a + x)^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} x +^{n} C_{2} a^{n - 2} x^{2} + \dots +^{n} C_{r} a^{n - r} x^{r} + \dots +^{n} C_{n} x^{n} = r = 0 \sum n^{n} C_{r} a^{n - r} x^{r}

where $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ are the binomial coefficients. Replacing $x$ with $- x$ :

(a - x)^{n} = r = 0 \sum n (- 1)^{r}^{n} C_{r} a^{n - r} x^{r}

Special case $a = 1$ :

(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + \dots +^{n} C_{n} x^{n}

(1 - x)^{n} =^{n} C_{0} -^{n} C_{1} x +^{n} C_{2} x^{2} - \dots + (- 1)^{n}^{n} C_{n} x^{n}

Key observations

There are exactly $(n + 1)$ terms in the expansion of $(a + x)^{n}$ .
In every term the sum of the powers of $a$ and $x$ equals $n$ .
The general term (the $(r + 1)$ -th term) is $T_{r + 1} =^{n} C_{r} a^{n - r} x^{r}$ .
The $p$ -th term from the end equals the $(n - p + 2)$ -th term from the beginning.
Coefficient of $x^{r}$ in $(a + x)^{n}$ is $^{n} C_{r} a^{n - r}$ .
$^{n} C_{x} =^{n} C_{y} \Rightarrow x = y$ or $x + y = n$ .
In $(a + x)^{n}$ and $(a - x)^{n}$ , the term containing $x^{r}$ is the $(r + 1)$ -th term.

Figure 1: Pascal's triangle - each row

n

lists the coefficients

^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}

(1 + x)^{n}

Solved Example 1

If the coefficients of the second, third and fourth terms in the expansion of

(1 + x)^{n}

are in A.P., show that

n = 7

Solution:

The coefficients of the 2nd, 3rd, 4th terms are $^{n} C_{1}$ , $^{n} C_{2}$ , $^{n} C_{3}$ . They are in A.P., so $2 \cdot^{n} C_{2} =^{n} C_{1} +^{n} C_{3}$ .

$\frac{2 n ( n - 1 )}{2} = n + \frac{n ( n - 1 ) ( n - 2 )}{6}$

Simplifying: $n^{2} - 9 n + 14 = 0 \Rightarrow (n - 2) (n - 7) = 0$ , giving $n = 2$ or $n = 7$ .

Since the symbol $^{n} C_{3}$ requires $n \geq 3$ , $n$ cannot be $2$ . Therefore $n = 7$ .

Solved Example 2

Find the (i) last digit, (ii) last two digits, and (iii) last three digits of

1 7^{256}

Solution:

$1 7^{256} = 28 9^{128} = (290 - 1)^{128}$ . Expanding:

$(290 - 1)^{128} =^{128} C_{0} (290)^{128} -^{128} C_{1} (290)^{127} + \dots +^{128} C_{126} (290)^{2} -^{128} C_{127} (290) + 1$

All terms with $(290)^{k}$ for $k \geq 3$ contain a factor of $1000$ , so mod $1000$ we get:

$1 7^{256} \equiv^{128} C_{2} (290)^{2} -^{128} C_{1} (290) + 1 (mod 1000)$

$= \frac{128 \cdot 127}{2} (290)^{2} - 128 (290) + 1 = 683, 527, 680 + 1 - 128 (290) + (terms in 1000)$ .

Reducing modulo $1000$ : last digit is $1$ , last two digits are $81$ , last three digits are $681$ .

Solved Example 3

If the binomial coefficients of the

(2 r + 4)

-th and

(r - 2)

-th terms in the expansion of

(a + b x)^{18}

are equal, find

r

Solution:

The $(k + 1)$ -th term has coefficient $^{18} C_{k}$ . So $^{18} C_{2 r + 3} =^{18} C_{r - 3}$ , which gives either

$2 r + 3 = r - 3 \Rightarrow r = - 6$ (not valid), or

$(2 r + 3) + (r - 3) = 18 \Rightarrow 3 r = 18 \Rightarrow r = 6$ .

Hence $r = 6$ is the only valid solution.

Solved Example 4

Find the coefficient of

x^{7}

(a x^{2} + \frac{1}{b x})^{11}

and the coefficient of

x^{- 7}

(a x - \frac{1}{b x ^{2}})^{11}

, and find the relation between

a

and

b

when these two coefficients are equal.

Solution:

General term of $(a x^{2} + \frac{1}{b x})^{11}$ is $T_{r + 1} =^{11} C_{r} (a x^{2})^{11 - r} (\frac{1}{b x})^{r} =^{11} C_{r} \frac{a ^{11 - r}}{b ^{r}} x^{22 - 3 r}$ .

For $x^{7}$ : $22 - 3 r = 7 \Rightarrow r = 5$ . Coefficient $=^{11} C_{5} \frac{a ^{6}}{b ^{5}}$ .

General term of $(a x - \frac{1}{b x ^{2}})^{11}$ is $T_{r + 1} = (- 1)^{r}^{11} C_{r} (a x)^{11 - r} (\frac{1}{b x ^{2}})^{r} = (- 1)^{r}^{11} C_{r} \frac{a ^{11 - r}}{b ^{r}} x^{11 - 3 r}$ .

For $x^{- 7}$ : $11 - 3 r = - 7 \Rightarrow r = 6$ . Coefficient $= (- 1)^{6}^{11} C_{6} \frac{a ^{5}}{b ^{6}} =^{11} C_{5} \frac{a ^{5}}{b ^{6}}$ (using $^{11} C_{6} =^{11} C_{5}$ ).

Equating: $\frac{a ^{6}}{b ^{5}} = \frac{a ^{5}}{b ^{6}} \Rightarrow a^{6} b^{6} = a^{5} b^{5} \Rightarrow ab = 1$ (assuming $a, b \neq = 0$ ).

3. Middle Term of a Binomial Expansion

The middle term of $(a + x)^{n}$ depends on whether $n$ is even or odd.

Case (a): $n$ is even. There is exactly one middle term,

T_{n /2 + 1} =^{n} C_{n /2} a^{n /2} x^{n /2}

.

Case (b): $n$ is odd. There are two middle terms,

T_{(n + 1) /2}

and

T_{(n + 3) /2}

, given by

^{n} C_{(n - 1) /2} a^{(n + 1) /2} x^{(n - 1) /2} and^{n} C_{(n + 1) /2} a^{(n - 1) /2} x^{(n + 1) /2} .

Solved Example 5

Find the middle term in the expansion of

(3 x - \frac{x ^{3}}{6})^{9}

Solution:

Since $n = 9$ is odd, there are two middle terms: $T_{5}$ and $T_{6}$ .

$T_{5} =^{9} C_{4} (3 x)^{5} (- \frac{x ^{3}}{6})^{4} = 126 \cdot 243 \cdot \frac{1}{1296} x^{17} = \frac{189}{8} x^{17}$ .

$T_{6} =^{9} C_{5} (3 x)^{4} (- \frac{x ^{3}}{6})^{5} = - \frac{21}{16} x^{19}$ .

Solved Example 6

Find the middle term in the expansion of

(\frac{a}{x} + b x)^{12}

Solution:

Since $n = 12$ is even, the middle term is $T_{7} = T_{6 + 1}$ .

$T_{7} =^{12} C_{6} (\frac{a}{x})^{6} (b x)^{6} =^{12} C_{6} a^{6} b^{6}$ .

4. Greatest Binomial Coefficient

In the expansion of $(1 + x)^{n}$ , the largest of the binomial coefficients $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ is called the greatest binomial coefficient.

If $n$ is even, the greatest coefficient is $^{n} C_{n /2}$ .
If $n$ is odd, the two coefficients $^{n} C_{(n - 1) /2}$ and $^{n} C_{(n + 1) /2}$ are equal and both are greatest.

5. Numerically Greatest Term

To find the term of largest absolute value in the expansion of $(1 + x)^{n}$ , examine the ratio of consecutive terms:

$\frac{T _{r + 1}}{T _{r}} = \frac{^{n} C _{r} x ^{r}}{^{n} C _{r - 1} x ^{r - 1}} = \frac{n - r + 1}{r} x$

Taking absolute values and requiring $∣ T_{r + 1} ∣ \geq ∣ T_{r} ∣$ :

$\frac{T _{r + 1}}{T _{r}} \geq 1 \Rightarrow \frac{( n - r + 1 ) ∣ x ∣}{r} \geq 1 \Rightarrow r \leq \frac{( n + 1 ) ∣ x ∣}{1 + ∣ x ∣}$

Procedure:

Compute $m = \frac{( n + 1 ) ∣ x ∣}{1 + ∣ x ∣}$ .
If $m$ is an integer, then $T_{m}$ and $T_{m + 1}$ are equal in magnitude and are the two greatest terms.
If $m$ is not an integer, then $T_{⌊ m ⌋ + 1}$ is the greatest term.

Solved Example 7

Find the greatest term in the expansion of

3 (1 + \frac{1}{3})^{20}

Solution:

The ratio $\frac{T _{r + 1}}{T _{r}} = \frac{^{20} C _{r}}{^{20} C _{r - 1}} \cdot \frac{1}{3} = \frac{21 - r}{r} \cdot \frac{1}{3}$ .

$T_{r + 1} \geq T_{r}$ requires $21 - r \geq r 3$ , i.e., $r \leq \frac{21}{3 + 1} = \frac{21 ( 3 - 1 )}{2} \approx 7.686$ .

So $T_{1} < T_{2} < \dots < T_{8} > T_{9} > T_{10}$ . The greatest term is $T_{8}$ .

$T_{8} = 3 \cdot^{20} C_{7} (\frac{1}{3})^{7} = 3 \cdot^{20} C_{7} \cdot \frac{1}{( 3 ) ^{6}} \cdot \frac{1}{3} =^{20} C_{7} \cdot \frac{1}{3 ^{3}} = \frac{77520}{27} \approx 2871.11$ .

Solved Example 8

Find numerically the greatest term in the expansion of

(3 - 5 x)^{11}

when

x = \frac{1}{5}

Solution:

Rewrite: $(3 - 5 x)^{11} = 3^{11} (1 - \frac{5 x}{3})^{11} = 3^{11} (1 - \frac{1}{3})^{11}$ (using $x = 1/5$ ).

For $(1 - \frac{1}{3})^{11}$ , compute $m = \frac{( 11 + 1 ) ∣ - 1/3 ∣}{1 + ∣ - 1/3 ∣} = \frac{12 \cdot \frac{1}{3}}{\frac{4}{3}} = 3$ .

Since $m = 3$ is an integer, $T_{3}$ and $T_{4}$ are both greatest.

$T_{3} = 3^{11}^{11} C_{2} (- \frac{1}{3})^{2} = 3^{11} \cdot 55 \cdot \frac{1}{9} = 55 \cdot 3^{9}$ .

$T_{4} = 3^{11}^{11} C_{3} (- \frac{1}{3})^{3} = 3^{11} \cdot 165 \cdot (- \frac{1}{27}) = - 55 \cdot 3^{9}$ . Both have magnitude $55 \cdot 3^{9}$ .

Common Mistakes to Avoid

Watch out

Confusing the " $r$ -th term" with the " $(r + 1)$ -th term": general term is $T_{r + 1} =^{n} C_{r} a^{n - r} x^{r}$ , indexed by $r$ starting at $0$ .
Forgetting alternating signs in $(a - x)^{n}$ : the $(r + 1)$ -th term carries $(- 1)^{r}$ .
Applying the middle-term formula for odd $n$ as if there were one middle term - there are two.
Mixing up the greatest binomial coefficient (a symmetry property) with the numerically greatest term (which depends on $x$ as well).
For numerically greatest term, forgetting to first pull out the factor to write the binomial in the standard $(1 + y)^{n}$ form.

Frequently Asked Questions

What is the binomial theorem in JEE and NEET Mathematics?

The binomial theorem is a formula for expanding $(a + x)^{n}$ as a finite series when $n$ is a positive integer: $(a + x)^{n} = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} x^{r}$ . It is a Class 11 chapter in the JEE Main and JEE Advanced syllabus. NEET does not include Mathematics, so this topic is relevant only for JEE (Main/Advanced) among the two exams.

How many terms are there in the expansion of $(a + x)^{n}$ ?

The expansion of $(a + x)^{n}$ contains exactly $n + 1$ terms, one for each value of $r$ from $0$ to $n$ . Each term has the form $T_{r + 1} =^{n} C_{r} a^{n - r} x^{r}$ , and the powers of $a$ and $x$ in every term sum to $n$ .

What is the general term of the binomial expansion?

The general term (also called the $(r + 1)$ -th term) in the expansion of $(a + x)^{n}$ is $T_{r + 1} =^{n} C_{r} a^{n - r} x^{r}$ . To find a specific term or a specific power of $x$ , set the exponent of $x$ equal to the required value and solve for $r$ .

How do I find the middle term of a binomial expansion?

If $n$ is even, there is one middle term, $T_{n /2 + 1}$ . If $n$ is odd, there are two middle terms, $T_{(n + 1) /2}$ and $T_{(n + 3) /2}$ . Substitute the appropriate value of $r$ into the general-term formula to get the middle term explicitly.

What is the difference between greatest binomial coefficient and numerically greatest term?

The greatest binomial coefficient is the largest value among $^{n} C_{0},^{n} C_{1}, \dots,^{n} C_{n}$ and depends only on $n$ (it is $^{n} C_{n /2}$ for even $n$ ). The numerically greatest term is the term with the largest absolute value in the actual expansion, and it depends on both $n$ and the value of $x$ .

How do I find the numerically greatest term in $(1 + x)^{n}$ ?

Compute $m = \frac{( n + 1 ) ∣ x ∣}{1 + ∣ x ∣}$ . If $m$ is an integer, both $T_{m}$ and $T_{m + 1}$ are numerically greatest. If $m$ is not an integer, $T_{⌊ m ⌋ + 1}$ is the greatest term. Always first rewrite the binomial in $(1 + y)^{n}$ form by factoring.

Is $(a + x)^{n}$ the same as $(x + a)^{n}$ in binomial expansion?

The two expressions are algebraically equal, but the expansions are written with different orderings of powers. $(a + x)^{n}$ starts with $a^{n}$ and ends with $x^{n}$ ; $(x + a)^{n}$ starts with $x^{n}$ and ends with $a^{n}$ . Same set of terms, opposite order.

What are common JEE-Main tricks tested from this concept?

Frequently tested: finding the coefficient of a specific power of $x$ ; identifying the middle term; equating two binomial coefficients using $^{n} C_{x} =^{n} C_{y} \Rightarrow x = y$ or $x + y = n$ ; finding last digits of large powers using expansion mod $10, 100, 1000$ ; A.P./G.P. conditions on consecutive term coefficients.

Previous year questions on Introduction to Binomial Expansion

19 questions from past papers, each with a step-by-step solution.

Show all 19 questions

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1. What is a Binomial Expression?

2. Binomial Theorem for a Positive Integer Index

Key observations

3. Middle Term of a Binomial Expansion

4. Greatest Binomial Coefficient

5. Numerically Greatest Term

Common Mistakes to Avoid

Frequently Asked Questions

What is the binomial theorem in JEE and NEET Mathematics?

How many terms are there in the expansion of (a+x)n?

What is the general term of the binomial expansion?

How do I find the middle term of a binomial expansion?

What is the difference between greatest binomial coefficient and numerically greatest term?

How do I find the numerically greatest term in (1+x)n?

Is (a+x)n the same as (x+a)n in binomial expansion?

What are common JEE-Main tricks tested from this concept?

Previous year questions on Introduction to Binomial Expansion

Ready to master Binomial Theorem And Mathematical Induction?

How many terms are there in the expansion of $(a + x)^{n}$ ?

How do I find the numerically greatest term in $(1 + x)^{n}$ ?

Is $(a + x)^{n}$ the same as $(x + a)^{n}$ in binomial expansion?