Binomial Theorem and its Simple Applications Formula Sheet — JEE Main Mathematics

Q: What are the most important Binomial Theorem and its Simple Applications formulas for JEE Main?

This Binomial Theorem and its Simple Applications formula sheet covers all the high-yield Mathematics formulas, definitions and theorems you need for JEE Main, across Binomial theorem for positive integral index, General term, Middle term, Simple applications — each shown with the key result and, where useful, a worked example.

Q: Is this Binomial Theorem and its Simple Applications formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

Q: How should I revise Binomial Theorem and its Simple Applications formulas?

Blurt the Binomial Theorem and its Simple Applications formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Every key Binomial Theorem and its Simple Applications formula, definition and theorem for JEE Main Mathematics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 4 sub-topics

Binomial theorem for positive integral index
General term
Middle term
Simple applications

Binomial Theorem — Statement & Pascal's Triangle

Binomial theorem (positive integral index):

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

, an expansion with

n + 1

terms. The powers of a fall from n to

0

while those of b rise from

0

to n, so the two exponents always add to n.

The binomial expansion formula, Pascal's triangle of coefficients, and the special cases (1+x)^n and (1-x)^n — The expansion, Pascal's triangle, and the special cases.

What to remember

▸There are $n + 1$ terms; the coefficients are the binomial coefficients $(r n)$ .
▸Pascal's triangle: each entry is the sum of the two above ( $(r n) + (r - 1 n) = (r n + 1)$ ).
▸ $(1 + x)^{n} = \sum (r n) x^{r}$ ; $(1 - x)^{n} = \sum (- 1)^{r} (r n) x^{r}$ .
▸Put $x = 1$ : the coefficients sum to $\sum (r n) = 2^{n}$ .

★ Remember · Exponents always add to n

In every term

(r n) a^{n - r} b^{r}

the powers of a and b sum to n. This is the quickest sanity check on any term you write down.

🎯 Exam · Symmetry of the coefficients

(r n) = (n - r n)

, so the coefficients read the same forwards and backwards — Pascal's rows are symmetric. The first and last coefficients are always

1

⚠️ Watch out · Mind the sign in (a−b)ⁿ

(a - b)^{n} = \sum (- 1)^{r} (r n) a^{n - r} b^{r}

— the terms with

odd

r are negative. Treat

- b

as a single quantity rather than dropping the sign halfway through.

🚫 Examiner Trap · Examiner traps

(1)

(a + b)^{n}

has

n + 1

terms (not n); exponents of a and b always SUM to n. (2)

(a - b)^{n}

— alternate signs

(- 1)^{r}

; only the odd-power terms flip vs

(a + b)^{n}

. (3) Pascal entry

=

sum of the two directly above; row n gives

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

. (4) Putting

x = 1

(1 + x)^{n}

gives

\sum^{n} C_{r} = 2^{n}

— a fast way to derive coefficient sums.

General Term & Middle Term

General term: The

(r + 1)

-th term of

(a + b)^{n}

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

, with

r = 0, 1, \dots, n

. Almost every binomial problem reduces to choosing the right value of r in this one formula.

Middle term of (a+b)ⁿ

▸n $even$ : one middle term, the $(\frac{n}{2} + 1)$ -th.
▸n $odd$ : two middle terms, the $\frac{n + 1}{2}$ -th and $\frac{n + 3}{2}$ -th.
▸The $(r + 1)$ -th term from the $end$ = the $(n - r + 1)$ -th from the start.
▸Use $T_{r + 1}$ for 'coefficient of', 'term containing', and 'middle term' questions alike.

★ Remember · The standard method

Write

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

, collect the net power of x, set it to the value you want, solve for r, then substitute r back to read off the term or coefficient.

⚠️ Watch out · (r+1)-th term, not r-th

T_{r + 1}

uses

(r n)

, so the term number is one more than the index r. The 5th term has

r = 4

, not

r = 5

— a constant source of off-by-one errors.

🚫 Examiner Trap · Examiner traps

(1) The general term is

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

— note it's the

(r + 1)

-th term, so r runs

0 \dots n

(off-by-one trap). (2) n EVEN

\Rightarrow

ONE middle term

(\frac{n}{2} + 1)

-th; n ODD

\Rightarrow

TWO middle terms. (3) The r-th term from the END is the

(n - r + 2)

-th from the start. (4) Almost every 'find the term' question is just solving for the right r.

Finding a Specific Term

★ Remember · Set the power of x

(a x^{p} + \frac{b}{x ^{q}})^{n}

the general term has net power of x equal to

p (n - r) - q r

. Set this

= 0

for the

term independent of x

(constant term), or

= k

for the coefficient of

x^{k}

, then solve for r.

Reading off the answer

▸Solve the power equation for r; it must be a whole number with $0 \leq r \leq n$ .
▸If r is not such an integer, $no$ such term exists.
▸Substitute r into $(r n) a^{n - r} b^{r}$ to get the term/coefficient.
▸For $(x + \frac{1}{x})^{n}$ : power $= n - 2 r$ , so the constant term needs $r = \frac{n}{2}$ (only if n even).

🎯 Exam · Ratio of consecutive coefficients

\frac{T _{r + 1}}{T _{r}} = \frac{n - r + 1}{r} \cdot \frac{b}{a}

. This relates one coefficient to the next, and is the engine behind both 'coefficients in A.P./G.P.' problems and the greatest-term method.

⚠️ Watch out · Coefficient vs term

The 'coefficient of

x^{k}

' is just the numerical part

(r n) a^{n - r} b^{r}

once r is fixed — do not include the

x^{k}

itself. The 'term' includes the power of x.

🚫 Examiner Trap · Examiner traps

(1) Constant term: set the NET power of x in

T_{r + 1}

0

and solve for r. (2) For

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

the net power is

n - 2 r

, so the constant term needs

r = \frac{n}{2}

— exists ONLY if n is even. (3) If the required r is NOT a whole number (

0 \leq r \leq n

), there is NO such term. (4) Use

\frac{T _{r + 1}}{T _{r}} = \frac{n - r + 1}{r} \cdot \frac{b}{a}

to relate consecutive coefficients.

Greatest Term & Greatest Coefficient

★ Remember · Numerically greatest term

Form

\frac{T _{r + 1}}{T _{r}} = \frac{n - r + 1}{r} \frac{b}{a}

. Let

m = \frac{( n + 1 ) ∣ b / a ∣}{1 + ∣ b / a ∣}

: if m is an integer,

T_{m}

and

T_{m + 1}

are equal and greatest; otherwise the greatest term is

T_{⌊ m ⌋ + 1}

Greatest binomial coefficient

▸The coefficients $(r n)$ are largest in the middle of the row.
▸n even: the single greatest is $(n /2 n)$ .
▸n odd: two equal greatest, $(( n - 1 ) /2 n) = (( n + 1 ) /2 n)$ .
▸They increase up to the middle and then decrease symmetrically.

⚠️ Watch out · Greatest term ≠ greatest coefficient

The greatest

coefficient

depends only on n (it is the middle

(r n)

). The greatest

term

depends on the actual values of a and b too — don't substitute one question's answer for the other.

🎯 Exam · Always plug in the numbers

For a numerical binomial (e.g. the greatest term in

(2 + 3 x)^{9}

x = \frac{1}{2}

), compute

∣ b / a ∣

with the actual values

before

finding m — the position of the greatest term shifts with x.

🚫 Examiner Trap · Examiner traps

(1) Don't confuse the numerically GREATEST TERM (depends on a,b,x) with the greatest COEFFICIENT (just

^{n} C_{r}

, independent of x). (2) Greatest term: compute

m = \frac{( n + 1 ) ∣ b / a ∣}{1 + ∣ b / a ∣}

; if integer,

T_{m} = T_{m + 1}

are both greatest, else

T_{⌊ m ⌋ + 1}

. (3) Greatest coefficient is at the MIDDLE:

r = \frac{n}{2}

(n even, one value) or two equal for n odd. (4) Use the ratio test

\frac{T _{r + 1}}{T _{r}} \geq 1

Binomial Coefficient Identities

Coefficient identities

(C_{r} =^{n} C_{r})

Identity	Formula
Sum of all	$C_{0} + C_{1} + \dots + C_{n} = 2^{n}$
Odd $=$ even halves	$C_{0} + C_{2} + \dots = C_{1} + C_{3} + \dots = 2^{n - 1}$
Weighted sum	$C_{1} + 2 C_{2} + \dots + n C_{n} = n 2^{n - 1}$
Alternating	$C_{0} - C_{1} + C_{2} - \dots = 0$ ( $n \geq 1$ )
Sum of squares	$C_{0}^{2} + C_{1}^{2} + \dots + C_{n}^{2} =^{2 n} C_{n}$
Vandermonde	$\sum_{k}^{m} C_{k}^{n} C_{p - k} =^{m + n} C_{p}$
Pascal	$^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
Absorption	$r^{n} C_{r} = n^{n - 1} C_{r - 1}$

Sums (write Cᵣ = C(n,r))

▸ $C_{0} + C_{1} + \dots + C_{n} = 2^{n}$ .
▸ $C_{0} + C_{2} + \dots = C_{1} + C_{3} + \dots = 2^{n - 1}$ .
▸ $C_{1} + 2 C_{2} + \dots + n C_{n} = n 2^{n - 1}$ .
▸ $C_{0}^{2} + C_{1}^{2} + \dots + C_{n}^{2} = (n 2 n)$ .

★ Remember · How they are proved

Substitute

x = 1

x = - 1

(1 + x)^{n}

for the plain sums;

differentiate

then put

x = 1

for the weighted sums

\sum r C_{r}

; and

compare coefficients

(1 + x)^{m} (1 + x)^{n} = (1 + x)^{m + n}

for products like

\sum C_{r}^{2}

(Vandermonde).

🎯 Exam · Vandermonde & absorption

Vandermonde:

\sum_{k} (k m) (p - k n) = (p m + n)

(with

m = n = p

it gives

\sum C_{r}^{2} = (n 2 n)

). Absorption:

r (r n) = n (r - 1 n - 1)

pulls a factor of r inside.

⚠️ Watch out · Differentiate, then substitute

For a sum like

\sum r^{2} C_{r}

, differentiate

(1 + x)^{n}

(sometimes twice, or multiply by x first)

before

putting

x = 1

. Substituting first loses the r-weighting you need.

🚫 Examiner Trap · Examiner traps

(1) Derive identities by SUBSTITUTING

x = 1

x = - 1

(1 + x)^{n}

, or by differentiating/integrating. (2)

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

; odd-indexed sum

=

even-indexed sum

= 2^{n - 1}

; alternating sum

C_{0} - C_{1} + \dots = 0

. (3) Weighted sum

\sum r C_{r} = n 2^{n - 1}

(differentiate). (4) Sum of squares

\sum C_{r}^{2} =^{2 n} C_{n}

(Vandermonde) — not

(2^{n})

Applications — Divisibility, Remainders & Parts

★ Remember · Expand around a multiple

Write the base as (multiple

\pm 1

8^{n} = (1 + 7)^{n} = 1 + 7 n + (2 n) 7^{2} + \dots

. Everything from

7^{2}

on is divisible by

49

, so

8^{n} \equiv 1 + 7 n (mod 49)

— the early low-power terms decide a remainder or last digit.

🎯 Exam · Integer & fractional parts

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

with

0 \leq f < 1

, add the conjugate

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

(small): the irrational parts cancel, so

(a + b)^{n} + (a - b)^{n}

is an

integer

and

f + f^{'}

is an integer (usually

1

★ Remember · Pin down f with the product

Since

(I + f) f^{'} = (a + b)^{n} (a - b)^{n} = (a^{2} - b)^{n}

, you can evaluate f exactly once you know f'. This is the standard route for 'find the fractional part' problems.

⚠️ Watch out · Check that the conjugate is small

The trick needs

∣ a - b ∣ < 1

so that

f^{'} = (a - b)^{n} \in (0, 1)

. If

a - b

is negative, watch the sign of f' — it may be

f + f^{'} = 1

rather than

0

, depending on the parity of n.

🚫 Examiner Trap · Examiner traps

(1) For divisibility, write the base as (multiple

\pm 1

) and expand: e.g.

8^{n} = (1 + 7)^{n}

, terms from

7^{2}

on are divisible by

49

. (2) The REMAINDER comes from the few low-power terms — don't expand fully. (3) Integer/fractional parts: for

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

, add the conjugate

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

; the irrational parts cancel leaving an integer, so

f + f^{'}

is an integer. (4) Then

(I + f) f^{'} = (a^{2} - b)^{n}

pins down f.

Binomial for Any Index & Multinomial

Binomial series (any index): For real n and

∣ x ∣ < 1

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

— an

infinite

series when n is negative or fractional (no longer

n + 1

terms).

Useful expansions & approximations

▸Small x: $(1 + x)^{n} \approx 1 + n x$ (first-order approximation).
▸ $(1 + x)^{- 1} = 1 - x + x^{2} - x^{3} + \dots$ .
▸ $(1 - x)^{- 1} = 1 + x + x^{2} + \dots$ ; $(1 - x)^{- 2} = 1 + 2 x + 3 x^{2} + \dots$ .
▸Always check the validity condition $∣ x ∣ < 1$ before using these.

🎯 Exam · Multinomial theorem

(x_{1} + \dots + x_{k})^{n} = \sum \frac{n !}{a _{1} ! \dots a _{k} !} x_{1}^{a_{1}} \dots x_{k}^{a_{k}}

over

a_{1} + \dots + a_{k} = n

. The number of distinct terms is

(k - 1 n + k - 1)

, and each coefficient is the multinomial coefficient

\frac{n !}{a _{1} ! \dots a _{k} !}

⚠️ Watch out · Positive-integer vs general index

The finite expansion (with

(r n)

and

n + 1

terms) is only for a

positive integer

n. For other n you must use the infinite series above, and only when

∣ x ∣ < 1

🚫 Examiner Trap · Examiner traps

(1) The general-index series

(1 + x)^{n}

(n real) is valid ONLY for

∣ x ∣ < 1

and is an INFINITE series (no

n + 1

terms, no

^{n} C_{r}

with factorials). (2) Small-x approximations:

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

;

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

. (3) Multinomial: distinct terms in

(x_{1} + \dots + x_{k})^{n}

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

. (4) The multinomial coefficient is

\frac{n !}{a _{1} ! \dots a _{k} !}

with

\sum a_{i} = n

More JEE Main Mathematics formula sheets

Frequently Asked Questions

What are the most important Binomial Theorem and its Simple Applications formulas for JEE Main?

This Binomial Theorem and its Simple Applications formula sheet covers all the high-yield Mathematics formulas, definitions and theorems you need for JEE Main, across Binomial theorem for positive integral index, General term, Middle term, Simple applications — each shown with the key result and, where useful, a worked example.

Is this Binomial Theorem and its Simple Applications formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Binomial Theorem and its Simple Applications formulas?

Blurt the Binomial Theorem and its Simple Applications formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Also useful: all formula sheets · JEE Main previous-year papers · most important chapters.