JEE Mains Maths · Binomial Theorem

Rational and Integral Terms

Expansions of sums of surds: counting or adding the terms that are rational or integral, and finding the integral part of a power of a surd by pairing it with its conjugate.

All Binomial Theorem notes JEE Mains Maths strategy

Why this matters

Ten PYQs, four of them numerical answer. Eight count or add rational terms, which is a divisibility condition on r; two need the integral part of a number like (7 + 4√3) to a power, where the conjugate does the work. Two ideas cover the page.

Concept 1 of 2: Counting and adding rational terms

\left(p^{1/a}+q^{1/b}\right)^n

the general term is

\binom nr p^{(n-r)/a}q^{r/b}

. It is rational exactly when

a

divides

n-r

and

b

divides

r

. So the rational terms are the

r

in an arithmetic progression; count them, or add those few terms. Irrational terms are the total

n+1

minus the rational ones.

Definition

Rational: $a\mid(n-r)$ and $b\mid r$ .
If $a\mid n$ , the condition is $\operatorname{lcm}(a,b)\mid r$ .
Number of multiples of $m$ in $[0,n]$ : $\left\lfloor\frac nm\right\rfloor+1$ .
Irrational terms $=n+1-$ rational terms.

Rational-term condition

\binom nr p^{\frac{n-r}{a}}q^{\frac rb}\ \text{is rational exactly when}\ a\mid(n-r)\ \text{and}\ b\mid r

Worked example

How many rational terms are in

\left(3^{1/2}+2^{1/3}\right)^{12}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q70Moderate

Example 1 · Binomial Theorem · Rational and Integral Terms

The sum of all rational terms in the expansion of

\left( 2^{\frac{1}{5}}+5^{\frac{1}{3}} \right)^{15}

is equal to:

Both exponents must be whole

Checking only

b\mid r

is not enough:

\frac{n-r}{a}

must be an integer too. When

a

does not divide

n

, the valid

r

are shifted, not multiples of

\operatorname{lcm}(a,b)

Drill 7 more on counting and adding rational terms

Concept 2 of 2: Integral parts through the conjugate

Let

x=(a+\sqrt b)^n

with conjugate

y=(a-\sqrt b)^n

. In

x+y

the surd terms cancel, so

x+y

is an integer. If

0<a-\sqrt b<1

, then

0<y<1

and

[x]=x+y-1

. If

-1<a-\sqrt b<0

y

has the sign of

(-1)^n

, and the integral part follows the same way.

Definition

$(a+\sqrt b)^n+(a-\sqrt b)^n=2\left[a^n+\binom n2a^{n-2}b+\dots\right]$ , an even integer.
$0<y<1\Rightarrow[x]=(x+y)-1$ , odd.
$-1<y<0\Rightarrow[x]=x+y$ , even.
$(7+4\sqrt3)(7-4\sqrt3)=1$ .

Conjugate pair

x=(a+\sqrt b)^n,\ y=(a-\sqrt b)^n:\quad x+y\in\mathbb Z,\ \ 0<y<1\Rightarrow[x]=x+y-1

Worked example

Is the integral part of

(2+\sqrt3)^{5}

odd or even?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 30 January 2023 · Q170Moderate

Example 2 · Binomial Theorem · Rational and Integral Terms

x = (8\sqrt{3}+ 13)^{13}

and

y = (7\sqrt{2}+ 9)^{9}

. If

\lbrack t\rbrack

denotes the greatest integer

\leq t

, then

A negative conjugate

When

a-\sqrt b

is negative,

y

changes sign with

n

. For odd

n

y

is negative and

[x]=x+y

exactly; subtracting 1 by habit gives the wrong parity.

Drill 1 more on integral parts through the conjugate

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)

Counting and adding rational terms
Rational-term condition
$\binom nr p^{\frac{n-r}{a}}q^{\frac rb}\ \text{is rational exactly when}\ a\mid(n-r)\ \text{and}\ b\mid r$
Integral parts through the conjugate
Conjugate pair
$x=(a+\sqrt b)^n,\ y=(a-\sqrt b)^n:\quad x+y\in\mathbb Z,\ \ 0<y<1\Rightarrow[x]=x+y-1$

Watch out for (2)

Both exponents must be whole
→ Counting and adding rational terms
A negative conjugate
→ Integral parts through the conjugate

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.

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