JEE Mains Maths · Binomial Theorem

Products and Multinomial Expansions

Finding a coefficient when the expression is a product of a polynomial and a binomial, a product of two binomials, or a power of a three-term bracket.

All Binomial Theorem notes JEE Mains Maths strategy

Why this matters

Nineteen PYQs, nine of them numerical answer. For a product, a coefficient is a short sum of products of coefficients. For a three-term bracket, either factor it into something simpler or use the multinomial term. Two ideas cover the page.

Concept 1 of 2: Coefficients in a product

The coefficient of

x^k

P(x)\,(1+x)^n

collects one term from each factor whose powers add to

k

. With a short polynomial in front, that is two or three terms. For

(1+x)^p(1-x)^q

, the coefficient of

x

p-q

and of

x^2

\frac{(p-q)^2-(p+q)}{2}

, which solves most such questions at once.

Definition

$[x^k]\,(a_0+a_1x+a_2x^2)(1+x)^n=a_0\binom nk+a_1\binom n{k-1}+a_2\binom n{k-2}$ .
$(1+x)^p(1-x)^q$ : $[x]=p-q$ , $[x^2]=\frac{(p-q)^2-(p+q)}{2}$ .
Rewrite first when it helps: $\left(1+\frac3x+\frac3{x^2}+\frac1{x^3}\right)^{5}=\left(1+\frac1x\right)^{15}$ .

Two factors

(1+x)^p(1-x)^q:\quad[x]=p-q,\quad[x^2]=\tfrac{(p-q)^2-(p+q)}{2}

Worked example

Find the coefficient of

x^3

(1+2x)(1+x)^{6}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q166Moderate

Example 1 · Binomial Theorem · Products and Multinomial Expansions

If the coefficients of

x

and

x^{2}

(1 + x)^{p}(1 - x)^{q}

are 4 and -5 respectively, then

2p + 3q

is equal to

Include every combination

With a quadratic in front, the coefficient of

x^k

has three contributions, from

x^0

x^1

and

x^2

. Dropping the last one is the usual slip.

Drill 9 more on coefficients in a product

Concept 2 of 2: Three-term brackets

First look for a factorisation:

(1-x)(1+x+x^2)=1-x^3

, so

(1-x)^{n+1}(1+x+x^2)^n=(1-x)(1-x^3)^n

, and only multiples of 3 survive. Otherwise use the multinomial term

\frac{n!}{a!\,b!\,c!}p^aq^br^c

with

a+b+c=n

, list the

(a,b,c)

that give the required power, and add.

Definition

$(p+q+r)^n=\sum\frac{n!}{a!\,b!\,c!}p^aq^br^c$ , $a+b+c=n$ .
$(1-x)(1+x+x^2)=1-x^3$ ; $1+x+x^2=(1-x)^2+3x$ .
Two conditions (on $a+b+c$ and on the power) leave one free index: list its values.

Multinomial term

(p+q+r)^n:\ \ \frac{n!}{a!\,b!\,c!}\,p^a\,q^b\,r^c,\quad a+b+c=n

Worked example

Find the coefficient of

x^4

(1+x+x^2)^{3}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q81Moderate

Example 2 · Binomial Theorem · Products and Multinomial Expansions

The coefficient of

x^{7}

\left( 1 - x + 2x^{3} \right)^{10}

List every triple

The two conditions usually allow several

(a,b,c)

. Write the free index's possible values in order before computing, so none is missed.

Drill 8 more on three-term brackets

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 (2)

Coefficients in a product
Two factors
$(1+x)^p(1-x)^q:\quad[x]=p-q,\quad[x^2]=\tfrac{(p-q)^2-(p+q)}{2}$
Three-term brackets
Multinomial term
$(p+q+r)^n:\ \ \frac{n!}{a!\,b!\,c!}\,p^a\,q^b\,r^c,\quad a+b+c=n$

Watch out for (2)

Include every combination
→ Coefficients in a product
List every triple
→ Three-term brackets

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 19 questions