Mathematics · Algebra · Chapter notes

Binomial Theorem · Class 11 Notes

Class 11 Mathematics notes on the Binomial Theorem: where the coefficients come from, Pascal's triangle, the general term, middle terms and the greatest coefficient, the greatest term, coefficient identities by substitution, differentiation and integration, remainders and divisibility, the multinomial expansion, and a JEE Advanced tier on integral and fractional parts.

Class 11JEE MainJEE AdvancedCBSE BoardsNCERTFree, no sign up

21Pages

10Diagrams

11Sections

Start reading Download PDF

In short

The binomial theorem is a counting result: expanding (a + b)^n, the coefficient of each term counts how many ways you can pick b from n brackets, which is why it is nCr. Everything else follows from the general term T(r+1) = nCr a^(n−r) b^r — the middle term, the greatest coefficient, and the identities you get by substituting values of a and b.

Contents

·How to Read This Set
1Where the Coefficients Come From the whole theorem is a count
2Pascal's Triangle the same numbers, built by adding
3The General Term one formula for every term
4Middle Terms and the Greatest Coefficient one peak or two
5The Greatest Term and why it is a different question
6Coefficient Identities substitute, differentiate, integrate
7Remainders and Divisibility the multiple-plus-one trick
8The Multinomial Expansion three or more terms in the bracket
9Integral and Fractional Parts JEE Advanced tier begins
10Advanced Worked Problems five problems, five techniques
11Beyond the Syllabus: Any Index
★Binomial Theorem · Fact Sheet

How to Read This Set

Sections 1 to 8 cover the full syllabus for JEE Main. Sections 9 and 10 are the JEE Advanced layer. There the answer is never one formula away. You have to build the route yourself.

The whole chapter in one line

A binomial coefficient counts the ways to choose.

Every result here follows from that. The expansion, Pascal's rule and the coefficient identities are all careful counting.

Think it through

What this chapter is worth

It appears in almost every JEE Main paper, usually as one or two questions.
The questions are short, so a clear method scores quickly. A vague one burns minutes.
Its identities come back later in probability and in sequences and series.

Trap alert

The habit that fixes this chapter

Write the general term first, every time. Then read the answer off it.
Almost every mistake comes from guessing a term number instead of solving for r.
Remember that r is the power of the second term, and the term is T_r+1.

Where the Coefficients Come From

Write out $(a+b)^4$ as four brackets multiplied together. To make one term, you pick one letter from each bracket. The coefficient of a term simply counts how many ways lead to it.

Picking b from exactly two of four brackets can be done in six ways. That count is the coefficient.

The binomial theorem, for a positive whole number n

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

where

^{n}C_r=\dfrac{n!}{r!\,(n-r)!}

Fact	Why
there are n + 1 terms	r runs from 0 up to n
the two powers always add to n	every term uses one letter from each of the n brackets
the power of a falls, the power of b rises	r counts the b's, so it goes up by one each term
the coefficients are symmetric	choosing r brackets for b is the same as choosing n - r for a

Pascal's Triangle

The same coefficients can be built without any factorials. Each number is the sum of the two just above it.

Row n gives the coefficients of (a+b) to the power n. The two 10s add to make the 20 below.

Pascal's rule

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

Choosing r + 1 from n + 1 things: either the last thing is chosen, or it is not. Those two cases give the two terms.

Property	Statement
Symmetry	$^{n}C_r={}^{n}C_{n-r}$
Edges	$^{n}C_0={}^{n}C_n=1$
Next to the edge	$^{n}C_1={}^{n}C_{n-1}=n$
Pull out a factor	$^{n}C_r=\dfrac{n}{r}\,{}^{n-1}C_{r-1}$
Ratio of neighbours	$\dfrac{^{n}C_r}{^{n}C_{r-1}}=\dfrac{n-r+1}{r}$

Think it through

Use the ratio of neighbours

The last row of that table does most of the work in sections 4 and 5.
It tells you whether the coefficients are still rising or have started to fall.
It needs no factorials at all, so it is fast in an exam.

The General Term

Seven terms of (a+b) to the sixth. The power of the second term is always r, one less than the term number.

The general term

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

The term that is r + 1 places from the start. The term r + 1 places from the end is

T_{n-r+1}

from the start.

Worked example 1

Find the coefficient of

x^3

(2x+3)^7

General term: $T_{r+1}={}^{7}C_r\,(2x)^{7-r}\,3^{r}$ .

The power of x is 7 - r. Set 7 - r = 3, so r = 4.

Coefficient $={}^{7}C_4\cdot 2^{3}\cdot 3^{4}=35\times 8\times 81$ .

Answer: 22680

Worked example 2

Find the term independent of x in

\left(x^2-\dfrac{1}{x}\right)^9

General term: $T_{r+1}={}^{9}C_r\,(x^2)^{9-r}\left(-\dfrac{1}{x}\right)^{r}={}^{9}C_r\,(-1)^r\,x^{18-3r}$ .

Independent of x means the power is zero. So 18 - 3r = 0, giving r = 6.

The term is $T_7={}^{9}C_6\,(-1)^6=84$ .

Answer: 84, the seventh term

Worked example 3

How many rational terms are there in

\left(\sqrt{2}+\sqrt[3]{3}\right)^{100}

General term: $T_{r+1}={}^{100}C_r\;2^{(100-r)/2}\;3^{r/3}$ .

Both powers must be whole numbers. So 100 - r is even, which means r is even. And r is a multiple of 3.

So r is a multiple of 6: 0, 6, 12, and so on up to 96.

Answer: 17 rational terms

Worked example 4

Find the 4th term from the end of

\left(2x-\dfrac{1}{x}\right)^{10}

There are 11 terms. Counting 4 from the end lands on term 11 - 4 + 1 = 8 from the start.

So use r = 7: $T_8={}^{10}C_7\,(2x)^{3}\left(-\dfrac{1}{x}\right)^{7}=120\times 8\times(-1)\,x^{3-7}$ .

Quick check: the k-th term from the end is the (n - k + 2)-th from the start.

Answer:

-\dfrac{960}{x^4}

Trap alert

Three slips worth avoiding

Calling it T_r. The formula gives T_r+1. Mixing these up shifts every answer by one term.
Dropping the sign. In $(x-y)^n$ the second term is -y, so a factor $(-1)^r$ appears.
Forgetting to raise the number as well as the x. In $(2x)^{7-r}$ the 2 is raised too.

Middle Terms and the Greatest Coefficient

The binomial coefficients rise to a peak in the middle, then fall again. Where that peak sits depends only on whether n is even or odd.

An even power has one tallest bar. An odd power has two equal tallest bars.

n is	Middle term(s)	Greatest binomial coefficient
even	one: $T_{n/2+1}$	$^{n}C_{n/2}$
odd	two: $T_{(n+1)/2}$ and $T_{(n+3)/2}$	$^{n}C_{(n-1)/2}={}^{n}C_{(n+1)/2}$

Worked example

Find the middle term of

\left(x+\dfrac{1}{x}\right)^{10}

n = 10 is even, so there is one middle term, $T_6$ , with r = 5.

$T_6={}^{10}C_5\,x^{5}\left(\dfrac{1}{x}\right)^{5}={}^{10}C_5$ .

Answer: 252, and it is independent of x

The Greatest Term

The greatest term is a different question from the greatest coefficient. The size of a term depends on the value of x. So the peak moves when x changes.

Keep moving right while the ratio of neighbouring terms stays above 1.

The method

Compare neighbours:

\dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\cdot\left|\dfrac{b}{a}\right|

Terms keep rising while this ratio is above 1. The last one before it drops is the greatest.
Shortcut: work out

m=\dfrac{(n+1)\,|y|}{1+|y|}

with

y=\dfrac{b}{a}

.
If m is not a whole number,

T_{[m]+1}

is greatest. If it is,

T_m

and

T_{m+1}

tie.

For (2 + 3x)^9 at x = 3/2, the greatest ...	is at	Depends on x?
binomial coefficient $^{9}C_r$	T5 and T6	no
numerical coefficient, the full number in front of x	T6 and T7, a tie	no
term, with x put in	T7 alone	yes

Trap alert

Read which greatest the question wants

Binomial coefficient means $^{n}C_r$ alone.
Numerical coefficient includes the constants from inside the bracket.
Term includes x as well. It is the only one of the three that changes with x.
One expansion can give three different answers, as the table shows.

Coefficient Identities

Start from one expansion, $(1+x)^n=\sum {}^{n}C_r\,x^r$ . Every identity comes from doing one of four things to both sides.

Height at x = 1, the value at x = -1, the slope at x = 1, and the area from 0 to 1.

Identity	Value	How you get it
$\sum\,{}^{n}C_r$	$2^n$	put x = 1
$\sum\,(-1)^r\,{}^{n}C_r$	0	put x = -1
sum of even-place coefficients	$2^{n-1}$	add the two results above
$\sum\,r\,{}^{n}C_r$	$n\,2^{n-1}$	differentiate, then put x = 1
$\sum\,r^2\,{}^{n}C_r$	$n(n+1)\,2^{n-2}$	differentiate, multiply by x, differentiate again
$\sum\,\dfrac{^{n}C_r}{r+1}$	$\dfrac{2^{n+1}-1}{n+1}$	integrate from 0 to 1
$\sum\,({}^{n}C_r)^2$	$^{2n}C_n$	coefficient of $x^n$ in $(1+x)^n(x+1)^n$
$\sum\,{}^{n}C_r\,{}^{n}C_{r+1}$	$^{2n}C_{n-1}$	same product, a different power of x
$\sum\,r\,({}^{n}C_r)^2$	$n\,{}^{2n-1}C_{n-1}$	pull out n, then the product again
$\sum_k\,{}^{m}C_k\,{}^{n}C_{r-k}$	$^{m+n}C_r$	Vandermonde: compare $x^r$ in $(1+x)^m(1+x)^n$

Think it through

Why the product trick works for the sum of squares

Write $(1+x)^n$ twice, but the second one backwards, as $(x+1)^n$ .
To get $x^n$ in the product, take $x^r$ from the first and $x^{n-r}$ from the second.
Those coefficients are $^{n}C_r$ and $^{n}C_{n-r}$ , which is $^{n}C_r$ again. So each pair gives a square.
The product is also $(1+x)^{2n}$ , whose $x^n$ coefficient is $^{2n}C_n$ .

Identities feeling like a list?

Watch each one appear on the curve.

Slide x, draw the tangent, shade the area, and see each identity come out as a height, a slope or an area.

Play this concept in the app

Remainders and Divisibility

To find a remainder, rewrite the base as a multiple of the divisor, plus or minus 1. Then expand. Every term except the last carries the divisor as a factor.

Powers of 5 cycle through four positions on a 13-hour clock. The binomial move gets there directly.

Worked example 1

Find the remainder when

5^{99}

is divided by 13.

$5^{2}=25=26-1$ , and 26 is a multiple of 13.

$5^{99}=5\cdot(26-1)^{49}$ . Every term of $(26-1)^{49}$ has a factor 26, except the last, $(-1)^{49}=-1$ .

So $5^{99}$ leaves $5\times(-1)=-5$ . Add 13 to make it positive.

Answer: 8

Worked example 2

Find the last two digits of

7^{100}

$7^{4}=2401=2400+1$ , and 2400 is a multiple of 100.

$7^{100}=(1+2400)^{25}$ . Every term after the first has a factor 2400.

Answer: 01

Worked example 3

Show that

3^{2n+2}-8n-9

is divisible by 64 for every positive whole number n.

$3^{2n+2}=9^{n+1}=(1+8)^{n+1}$ .

Expand: $1+(n+1)\cdot 8+{}^{n+1}C_2\cdot 8^{2}+{}^{n+1}C_3\cdot 8^{3}+\dots$

The first two terms give $8n+9$ , which cancels exactly. Every term left has $8^2=64$ in it.

Proved: the whole expression is a multiple of 64

Trap alert

A negative remainder is not the answer

The method often ends at a value such as -5. A remainder must lie between 0 and the divisor.
Add the divisor once: -5 + 13 = 8.

The Multinomial Expansion

With three or more terms in the bracket, the same choosing idea still works. Now you choose which letter to take from each bracket, out of three or more options.

Ten stars for the power, two bars to split them among three letters.

The two results

Coefficient of

a^{p}\,b^{q}\,c^{s}

(a+b+c)^n

with p + q + s = n:

\dfrac{n!}{p!\,q!\,s!}

Number of terms in

(x_1+x_2+\dots+x_k)^n

^{n+k-1}C_{k-1}

Worked example

Find the coefficient of

x^4

(1+x+x^2)^5

Take 1 from p brackets, x from q brackets and $x^2$ from s brackets. Then p + q + s = 5, and the power is q + 2s = 4.

The options are (s, q, p) = (0, 4, 1), (1, 2, 2) and (2, 0, 3).

They give $\dfrac{5!}{1!\,4!\,0!}+\dfrac{5!}{2!\,2!\,1!}+\dfrac{5!}{3!\,0!\,2!}=5+30+10$ .

Answer: 45

Think it through

Counting terms when powers can merge

The formula counts terms in separate letters, such as a, b and c.
In $(1+x+x^2)^5$ the letters are all powers of x, so many terms merge.
The powers run from 0 to 10, so there are only 11 distinct terms, not 21.

Tier 2

The JEE Advanced layer

Everything so far is complete for JEE Main. From here each problem needs a route you construct yourself. No single formula reaches the answer.

Integral and Fractional Parts

A number like $(\sqrt{3}+1)^6$ is irrational, yet questions ask for its whole-number part. The trick is to pair it with its conjugate, $(\sqrt{3}-1)^6$ , which is small.

The two fractional parts must add to exactly 1, because the total is a whole number.

Conjugates still feel like a trick?

Push the power up and watch the two fractions keep adding to 1.

Raise n one step at a time. The big number grows, the small partner shrinks, and f + f' stays fixed at exactly 1.

Play this concept in the app

The general argument

Let

(\sqrt{a}+b)^n=I+f

with I a whole number and 0 < f < 1.
Let

f'=(\sqrt{a}-b)^n

, chosen so that 0 < f' < 1.
Adding the two expansions cancels every odd power of

\sqrt{a}

. So I + f + f' is a whole number.
So f + f' is a whole number between 0 and 2. It must equal 1.

Think it through

When the partner is subtracted instead

If $f'$ is small but can be negative, use the difference $I+f-f'$ instead.
Then the even powers cancel, and $f-f'$ is a whole number in (-1, 1), so it is 0.
So f = f'. Pick the sum or the difference by which one cancels the irrational part.

Advanced Worked Problems

Each problem uses a different technique. None of them can be solved by recalling a formula.

Problem 1 · integration

Find

\sum_{r=0}^{n}\dfrac{(-1)^r\,{}^{n}C_r}{r+1}

What it demands: turning the coefficient sum into an integral, and choosing the limits 0 and 1

Start from $(1-x)^n=\sum (-1)^r\,{}^{n}C_r\,x^r$ .

Integrate both sides from 0 to 1. The right side becomes the sum we want, since $\int_0^1 x^r\,dx=\dfrac{1}{r+1}$ .

The left side is $\int_0^1 (1-x)^n\,dx=\dfrac{1}{n+1}$ .

Answer:

\dfrac{1}{n+1}

Problem 2 · hockey stick twice

Find the coefficient of

x^{50}

\sum_{k=0}^{1000}(k+1)\,x^{k}(1+x)^{1000-k}

What it demands: rewriting a weight as a count, swapping the order of summation, and applying the same identity twice

The term $x^k(1+x)^{1000-k}$ contributes $^{1000-k}C_{50-k}$ to $x^{50}$ . So we need $\sum_{k=0}^{50}(k+1)\,{}^{1000-k}C_{50-k}$ .

Write the weight k + 1 as a count: it is the number of j from 0 to k. Swap the order of summing.

For fixed j, $\sum_{k=j}^{50}{}^{1000-k}C_{50-k}={}^{1001-j}C_{50-j}$ , by the hockey stick identity.

Then $\sum_{j=0}^{50}{}^{1001-j}C_{50-j}={}^{1002}C_{50}$ , by the hockey stick again.

Answer:

^{1002}C_{50}

Problem 3 · truncation

Find the last three digits of

17^{256}

What it demands: choosing a base close to a round number, then deciding which terms can still matter

$17^{2}=289=290-1$ . So $17^{256}=(1-290)^{128}$ .

Work modulo 1000. Since $290^{3}$ ends in 000, only three terms survive.

$1-128(290)+{}^{128}C_2\,(290)^2=1-37120+8128\times 84100$ .

Modulo 1000 these are 1, 880 and 800. Their sum is 1681.

Answer: 681

Problem 4 · parity

Show that the integral part of

(5+2\sqrt{6})^n

is odd for every positive whole number n.

What it demands: a counterintuitive claim about an irrational number, settled by the conjugate and a parity argument

Let $(5+2\sqrt{6})^n=I+f$ . Let $f'=(5-2\sqrt{6})^n$ . Since $5-2\sqrt{6}\approx 0.1$ , we have 0 < f' < 1.

Adding cancels the odd powers of $\sqrt{6}$ : $I+f+f'=2\left[5^n+{}^{n}C_2\,5^{n-2}(24)+\dots\right]$ , an even number.

So f + f' = 1, and I = even - 1.

Proved: I is always odd

Problem 5 · optimisation

For which x > 0 is

T_6

the greatest term of

(1+x)^{10}

What it demands: turning 'greatest' into two simultaneous inequalities, then handling the equal cases

$T_6$ is greatest when $T_6\geq T_5$ and $T_6\geq T_7$ .

$\dfrac{T_6}{T_5}=\dfrac{^{10}C_5}{^{10}C_4}\,x=\dfrac{6}{5}x\geq 1$ gives $x\geq\dfrac{5}{6}$ .

$\dfrac{T_7}{T_6}=\dfrac{^{10}C_6}{^{10}C_5}\,x=\dfrac{5}{6}x\leq 1$ gives $x\leq\dfrac{6}{5}$ .

At the two ends there is a tie, with $T_5$ or $T_7$ .

Answer:

\dfrac{5}{6}\leq x\leq\dfrac{6}{5}

, strictly greatest inside the interval

Beyond the Syllabus: Any Index

As we read them, the current JEE Main and JEE Advanced syllabi cover only a positive whole number index. The expansion below still turns up in older question banks and other exams. It is kept short, so skip it if time is tight.

The series follows the curve closely near 0, then parts company once x moves past 1.

For any real n, valid only when |x| < 1

(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\dots

Expansion	Coefficient of $x^r$
$(1-x)^{-1}$	1
$(1-x)^{-2}$	r + 1
$(1-x)^{-3}$	$\dfrac{(r+1)(r+2)}{2}$
$(1-x)^{-n}$	$^{n+r-1}C_r$

Trap alert

Two conditions, both required

The first term must be 1. For $(2+x)^n$ , first write it as $2^n(1+\dfrac{x}{2})^n$ .
Then the variable part must lie between -1 and 1. Outside that range the series does not settle to any value.

★ Binomial Theorem · Fact Sheet

Every rule for revision day. Print this page alone.

THE THEOREM

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

n + 1 terms.
The powers add to n.

GENERAL TERM

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

r is the power of the SECOND term.
From the end: $T_{n-r+1}$ .

PASCAL

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

Symmetric rows.
Row n adds to $2^n$ .

MIDDLE TERMS

n even: one,

T_{n/2+1}

n odd: two, which tie
for greatest coefficient.

GREATEST TERM

m=\dfrac{(n+1)|y|}{1+|y|}

Not whole: $T_{[m]+1}$ .
Whole: $T_m$ and $T_{m+1}$ tie.

THREE GREATESTS

Binomial coeff, numerical coeff,

and term can all differ.
Only the term depends on x.

SUMS

\sum{}^{n}C_r=2^n

$\sum r\,{}^{n}C_r=n\,2^{n-1}$
even places = odd places.

SQUARES

\sum({}^{n}C_r)^2={}^{2n}C_n

Vandermonde: $\sum{}^{m}C_k{}^{n}C_{r-k}={}^{m+n}C_r$ .

CALCULUS

\sum\dfrac{^{n}C_r}{r+1}=\dfrac{2^{n+1}-1}{n+1}

Integrate 0 to 1.
Differentiate for r-weights.

REMAINDERS

Base = multiple of divisor, plus or minus 1.

Only the last term survives.
Fix a negative remainder.

MULTINOMIAL

Coeff

\dfrac{n!}{p!\,q!\,s!}

Terms: $^{n+k-1}C_{k-1}$
Merging powers means fewer.

INTEGRAL PART

Pair with the conjugate.

f + f' = 1 when f' is in (0,1).
$(5+2\sqrt{6})^n$ has odd I.

More mathematics notes

You have read it

Now play it.

Reading a chapter and understanding it are different things. In the app this chapter becomes a game you play, a short read, then practice that tests you at every step. A full chapter, understood, in 45 to 60 minutes. Free to start.

PlayA game built for the concept
ReadThe short NCERT explainer
PracticeTimed, until it sticks

Still stuck at 11pm? TarQPro reads the answer you got wrong and teaches the idea behind it, not just the right option.

Get the app See how a chapter works