JEE Mains Maths · Binomial Theorem

The General Term

Finding one coefficient or the term independent of x in a single binomial expansion: write the general term, set its power of x, and read off r.

All Binomial Theorem notes JEE Mains Maths strategy

Why this matters

Thirty-four PYQs, eighteen of them numerical answer, and 2023 alone has fourteen. Most ask for one coefficient or the constant term; some first simplify the bracket, and some equate a coefficient from two different expansions. Three ideas cover the page.

Concept 1 of 3: Setting the power of x

The

(r+1)

-th term of

(a+b)^n

T_{r+1}=\binom nr a^{n-r}b^r

. When

a

and

b

carry powers of

x

, collect the exponent of

x

as a linear expression in

r

, set it equal to the power asked for, and solve. A non-integer

r

means that term does not exist. Keep the signs and the numerical factors of

a

and

b

inside the coefficient.

Definition

$T_{r+1}=\binom nr a^{n-r}b^{r}$ , $r=0,1,\dots,n$ .
$a=px^{\alpha}$ , $b=qx^{\beta}$ : power of $x$ is $\alpha(n-r)+\beta r$ .
Term independent of $x$ : set the power to 0.
No integer $r$ in $[0,n]$ : the coefficient is 0.

General term

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

Worked example

Find the term independent of

x

\left(x^2-\frac{2}{x}\right)^{6}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q57Moderate

Example 1 · Binomial Theorem · The General Term

The coefficient of

x^{2}

in the expansion of

\left( 2x^{2}+\frac{1}{x} \right)^{10},x \neq 0

, is :

T with index r + 1

The term containing

b^r

is the

(r+1)

-th. 'The 7th term' means

r=6

; using

r=7

shifts every answer.

Drill 20 more on setting the power of x

Concept 2 of 3: Simplify the bracket first

Some brackets are designed to collapse.

\frac{x+1}{x^{2/3}-x^{1/3}+1}

x^{1/3}+1

by the sum of cubes, and

\frac{x-1}{x-\sqrt x}

1+x^{-1/2}

. Others carry a parameter: a term containing

x^{\log_2x}

becomes an equation in

t=\log_2x

, and a term such as

\binom{10}{5}x(1-x)

is maximised like any function.

Definition

$x+1=(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)$ .
$x-1=(\sqrt x-1)(\sqrt x+1)$ , $x-\sqrt x=\sqrt x(\sqrt x-1)$ .
$3^{\log_3u}=u$ : simplify such bases before expanding.
Maximum of $x(1-x)$ on $[0,1]$ is $\frac14$ .

A common collapse

\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-\sqrt x}=x^{1/3}-x^{-1/2}

Worked example

Find the term independent of

x

\left(\frac{x-1}{\sqrt x-1}-1\right)^{8}

for

x>1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 10 · Q84Moderate

Example 2 · Binomial Theorem · The General Term

The term independent of

x

in the expansion of

\left\lbrack \frac{x + 1}{x^{2/3}-x^{1/3}+ 1}-\frac{x - 1}{x -x^{1/2}} \right\rbrack^{10},x \neq 1

, is equal to

Check the numerator's sign

\frac{x-1}{x-\sqrt x}

collapses;

\frac{x+1}{x-\sqrt x}

does not. If the bracket will not simplify, re-read it before expanding a messy expression.

Drill 6 more on simplify the bracket first

Concept 3 of 3: Equating coefficients from two expansions

When the coefficient of one power in one expansion equals that of another power in a second, find each

r

separately, write both coefficients, and divide. Often the binomial coefficients are equal by symmetry,

\binom{n}{r}=\binom{n}{n-r}

, and cancel, leaving a relation such as

ab=1

Definition

Find $r$ in each expansion from its own power of $x$ .
$\binom nr=\binom n{n-r}$ : the two binomial coefficients often cancel.
Ratio of two terms of one expansion: $\frac{T_{s+1}}{T_{r+1}}$ keeps only the changed powers.

Symmetry of coefficients

\binom nr=\binom n{n-r}

Worked example

The coefficient of

x^4

\left(ax+\frac1x\right)^{6}

equals that of

x^{-2}

\left(x+\frac{a}{x}\right)^{6}

a>0

. Find

a

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q75Moderate

Example 3 · Binomial Theorem · The General Term

If the coefficient of

x^{7}

\left( ax -\frac{1}{bx^{2}} \right)^{13}

and the coefficient of

x^{- 5}

\left( ax +\frac{1}{bx^{2}} \right)^{13}

are equal, then

a^{4}b^{4}

is equal to:

Signs from the second term

\left(ax-\frac{1}{bx^2}\right)^{n}

the term carries

(-1)^r

. If the two coefficients have opposite signs, they cannot be equal for positive

a,b

— check the parity of each

r

Drill 5 more on equating coefficients from two expansions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Setting the power of x
General term
$T_{r+1}=\binom nr\,a^{\,n-r}\,b^{\,r}$
Simplify the bracket first
A common collapse
$\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-\sqrt x}=x^{1/3}-x^{-1/2}$
Equating coefficients from two expansions
Symmetry of coefficients
$\binom nr=\binom n{n-r}$

Watch out for (3)

T with index r + 1
→ Setting the power of x
Check the numerator's sign
→ Simplify the bracket first
Signs from the second term
→ Equating coefficients from two expansions

Test yourself on Binomial Theorem

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Binomial Theorem chapter test Drill all 34 questions