Class 11 Maths Notes

Binomial Theorem Class 11 Notes

Complete, exam-ready notes on the binomial theorem: the expansion of (a + b)^n for positive integral n, the general term T_{r+1}, middle term for even and odd n, properties of binomial coefficients including symmetry and sum identities, and applications to simplification and coefficient extraction — written for CBSE boards and JEE revision.

Class11SubjectMathematicsCoversCBSE · JEE

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What does the binomial theorem do?

It expands (a + b)^n into a sum of terms involving binomial coefficients, avoiding the tedious process of repeated multiplication.

Binomial Expansion for Positive Integral Index

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

Binomial theorem

The expansion of $(a+b)^n$ has (n + 1) terms. The powers of a decrease from n to 0 while the powers of b increase from 0 to n. Each term's coefficient is the corresponding binomial coefficient $^nC_r$ .

For (1 + x)^n, set a = 1 to get a simpler form:
$(1+x)^n = \sum_{r=0}^{n} {^nC_r} x^r$
.
The general term in the expansion of (a + b)^n is the (r + 1)-th term.
The sum of all binomial coefficients equals 2^n.

Pascal's Triangle

Pascal's triangle

A triangular array where each entry is the sum of the two entries directly above it. Row n (starting from row 0) gives the binomial coefficients $^nC_0, \; ^nC_1, \; \ldots, \; ^nC_n$ .

Row 0: 1. Row 1: 1 1. Row 2: 1 2 1. Row 3: 1 3 3 1.
Each row is symmetric: the first and last entries are 1, and interior entries are sums of the two above.
Pascal's identity
$^nC_r + ^nC_{r-1} = ^{n+1}C_r$
is the rule that builds the triangle.

General Term

T_{r+1} = {^nC_r} \, a^{n-r} \, b^r

General (r + 1)-th term

The (r + 1)-th term in the expansion of $(a+b)^n$ is obtained by substituting into the general term formula. For example, the 4th term (r = 3) in $(2x + 3)^5$ is $^5C_3 (2x)^2 (3)^3$ .

Term index vs power

The (r + 1)-th term involves r as the exponent of b and (n − r) as the exponent of a. Always remember: r counts from 0, so the first term is T₁ with r = 0.

Middle Term

When n is even (say n = 2m), there is a single middle term: the (m + 1)-th term, i.e. T_{m+1}.
When n is odd (say n = 2m + 1), there are two middle terms: the (m + 1)-th and (m + 2)-th terms.

n \text{ even } (n=2m): \; T_{m+1} = {^{2m}C_m} \, a^m \, b^m

Middle term (n even)

The middle term carries the largest binomial coefficient in the expansion. For (1 + x)^n with even n, the middle term is the numerically largest term when all binomial coefficients are positive.

Properties of Binomial Coefficients

Symmetry:
$^nC_r = ^nC_{n-r}$
— the r-th and (n − r)-th coefficients are equal.
Sum of coefficients:
$\sum_{r=0}^{n} {^nC_r} = 2^n$
(set a = b = 1).
Alternating sum:
$\sum_{r=0}^{n} (-1)^r {^nC_r} = 0$
(set a = 1, b = −1).
Weighted sum:
$\sum_{r=0}^{n} r \, {^nC_r} = n \cdot 2^{n-1}$
(differentiate and set x = 1).

Even n has one middle term, odd n has two

Don't confuse the two cases. For n = 6, the single middle term is the 4th term (r = 3). For n = 7, the two middle terms are the 4th (r = 3) and 5th (r = 4).

Applications and Simplifications

Approximate (1 + x)^n ≈ 1 + nx for small x (first-order binomial approximation, useful in physics).
Coefficient extraction: the coefficient of x^r in (1 + x)^n is simply
$^nC_r$
.
For (a + bx)^n, the coefficient of x^r is
$^nC_r \, a^{n-r} \, b^r$
.
Sums involving binomial coefficients can often be evaluated by choosing specific values of x in the expansion.

Solved Examples

Example: Find the 5th term in the expansion of $(3x - 2)^8$ .

Solution: The 5th term has r = 4. $T_5 = {^8C_4} (3x)^4 (-2)^4 = 70 \times 81x^4 \times 16 = 90720 \, x^4$ .

Example: Find the middle term in the expansion of $(x + \frac{1}{x})^{10}$ .

Solution: n = 10 (even), so the single middle term is T₆ with r = 5. $T_6 = {^{10}C_5} \, x^5 \, x^{-5} = 252$ . The middle term is the constant 252.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Binomial expansion

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

General term

T_{r+1} = {^nC_r} \, a^{n-r} \, b^r

Middle term (n even)

T_{m+1} = {^{2m}C_m} \, a^m \, b^m

Symmetry

^nC_r = ^nC_{n-r}

Sum of coefficients

\sum_{r=0}^{n} {^nC_r} = 2^n

Alternating sum

\sum_{r=0}^{n} (-1)^r {^nC_r} = 0

(1 + x)^n expansion

(1+x)^n = \sum_{r=0}^{n} {^nC_r} x^r

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

The expansion of (a + b)^n has (n + 1) terms; the general term is T_{r+1} = nCr a^{n−r} b^r.
For even n, one middle term at position (n/2 + 1); for odd n, two middle terms.
Set a = b = 1 for sum of coefficients = 2^n; set a = 1, b = −1 for alternating sum = 0.
Binomial coefficients are symmetric: nCr = nC(n − r) — use the smaller value to compute.
Approximation (1 + x)^n ≈ 1 + nx works when |x| is small.
Coefficient of x^r in (1 + x)^n is nCr directly.

FAQ

Common questions

What is the binomial theorem?

It states that (a + b)^n = Σ nCr a^{n−r} b^r for positive integral n, expanding a power of a binomial into a sum of terms with binomial coefficients.

How do you find the middle term?

If n is even (n = 2m), the single middle term is T_{m+1}. If n is odd (n = 2m+1), the two middle terms are T_{m+1} and T_{m+2}.

What is Pascal's triangle?

A triangular array where each entry is the sum of the two entries above it. Row n gives the binomial coefficients nC0, nC1, ..., nCn for the expansion of (a + b)^n.

Why does the sum of binomial coefficients equal 2^n?

Setting a = b = 1 in the expansion gives (1 + 1)^n = Σ nCr = 2^n. Each term contributes its coefficient with no powers of a or b, summing all coefficients.

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