Chapter 5 of 16

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JEE Main Mathematics★ ~1 question (4 marks) in JEE Main, usually a general-term computation that is fully scripted once the method is fixed, making it among the most reliable marks in the paper

Binomial Theorem

Every binomial coefficient is a count of choices, which is why substitution answers questions that summation cannot. Set x to 1 and twenty-one numbers add themselves.

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By the end of this chapter you'll be able to…

1Derive the binomial coefficients as counts of choices among the n brackets, and use that reading to explain their symmetry
2Write the general term with the correct index and use it to find a specified coefficient, a term independent of x, or the middle term or terms
3Evaluate families of coefficient sums by substituting values into the expression, including the even and odd splits
4Recognise when an identity calls for differentiation, integration or a product of two expansions
5Distinguish the greatest coefficient from the greatest term and locate both from the ratio of consecutive terms
6Apply the expansion to divisibility, remainders, last digits, surd conjugate pairs and small-x approximation

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Why this chapter matters in JEE Main

Adding the twenty-one coefficients of (1+x) to the twentieth looks like arithmetic, but nobody adds them: put x equal to 1 and the expansion becomes the sum, giving 2 to the twentieth. Now try the same on (2+3x) to the twentieth. The same twenty-one combinations appear in it, so the reflex answers 2 to the twentieth again, and it is wrong by a factor of about ninety million, because the answer is 5 to the twentieth. The combination was never the coefficient: the coefficient of x to the r is C(20,r) times 2 to the 20-r times 3 to the r, and the powers carry nearly all the size. Substitute into the expression, never into a remembered shape. Substitution works at all because the theorem is a counting statement, with each coefficient counting which brackets supplied the second letter, and that single reading rebuilds every identity in the chapter instead of storing it.

Before you start — revise these

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Combinations and the identity C(n,r) = C(n,n-r), from Permutations and Combinations

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Laws of indices, including negative and fractional exponents

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Basic modular arithmetic: what a remainder is and how to correct a negative one

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Elementary differentiation and integration of powers of x, for the identity derivations

Binomial Theorem

Add up all twenty-one coefficients in the expansion of $(1 + x)^{20}$ .

Nobody adds them. Put $x = 1$ and the expansion becomes the sum you wanted.

$\sum_{r = 0}^{20} (r 20) = (1 + 1)^{20} = 2^{20} = 1, 048, 576$

Now do the same for $(2 + 3 x)^{20}$ . The same twenty-one combinations $(r 20)$ appear in it, so the reflex answers $2^{20}$ again.

$Actual answer: (2 + 3)^{20} = 5^{20} = 95, 367, 431, 640, 625$

Wrong by a factor of about ninety million.

The reason is that $(r 20)$ was never the coefficient. The coefficient of $x^{r}$ in $(2 + 3 x)^{20}$ is $(r 20) 2^{20 - r} 3^{r}$ , and the powers of $2$ and $3$ carry almost all of the size.

Substitution	What it delivers
$x = 1$	sum of all coefficients
$x = - 1$	alternating sum, which is $0$ whenever the binomial is $(1 + x)^{n}$
average the two	sum of the even-indexed coefficients

Substitute into the expression, never into a remembered shape.

Substitution works at all because the theorem is a counting statement. Multiplying out $n$ identical brackets means taking either $a$ or $b$ from each. A term with $r$ copies of $b$ arises exactly once for every way of choosing which $r$ brackets supplied the $b$ .

$coefficient of a^{n - r} b^{r} = number of ways to choose r brackets from n = (r n)$

That is why the coefficients are combinations, why they are symmetric, and why every identity in this chapter can be rebuilt rather than recalled.

1. The Theorem

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

Feature	Consequence
Number of terms	$n + 1$ , not $n$
Powers of $a$	fall from $n$ to $0$
Powers of $b$	rise from $0$ to $n$
Sum of the two powers	always $n$ , in every term
Coefficients	symmetric, since $(r n) = (n - r n)$

The last row is a free error check: if your expansion is not palindromic in its coefficients, something has gone wrong.

Two special forms are worth writing out, because most questions arrive in one of them.

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

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

The signs in the second alternate because $(- x)^{r}$ carries $(- 1)^{r}$ . Nothing else changes, and in particular the magnitudes of the coefficients are identical.

Trap. The theorem stated here is for a positive integral index $n$ . That is what makes the expansion finite, with exactly $n + 1$ terms, and what makes every coefficient a genuine count of choices.

Illustration 1

Find the coefficient of $a^{2} b^{3}$ in $(a + b)^{5}$ by counting, and confirm it against the formula.

Five brackets, and a term $a^{2} b^{3}$ needs the $b$ taken from exactly three of them. Choose which three.

$(3 5) = 10$

Listing confirms it: the three $b$ -brackets can be ${1, 2, 3}$ , ${1, 2, 4}$ , ${1, 2, 5}$ , ${1, 3, 4}$ , ${1, 3, 5}$ , ${1, 4, 5}$ , ${2, 3, 4}$ , ${2, 3, 5}$ , ${2, 4, 5}$ , ${3, 4, 5}$ .

Each of those ten selections produces the identical product $a^{2} b^{3}$ , so they collect into the single term $10 a^{2} b^{3}$ .

Note that $(3 5) = (2 5) = 10$ says the same thing twice: choosing which three brackets give $b$ is choosing which two give $a$ .

2. The General Term

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

Trap. The subscript is $r + 1$ , not $r$ . The first term has $r = 0$ . So the $4$ th term of $(a + b)^{10}$ uses $r = 3$ and is $(3 10) a^{7} b^{3} = 120 a^{7} b^{3}$ , not $(4 10) a^{6} b^{4} = 210 a^{6} b^{4}$ .

The working method for every "find the coefficient of" question is the same three steps.

$write T_{r + 1} \Rightarrow collect the power of x \Rightarrow set that power equal to the target and solve for r$

If $r$ comes out non-integral or outside $0 \leq r \leq n$ , that power does not occur and the coefficient is $0$ . That is a legitimate answer, not a mistake.

Illustration 2

Find the $5$ th term of $(2 x + \frac{1}{x})^{9}$ .

Fifth term means $r = 4$ .

$T_{5} = (4 9) (2 x)^{9 - 4} (\frac{1}{x})^{4} = 126 \times 32 x^{5} \times x^{- 4}$

$T_{5} = 4032 x$

Two things had to be carried that a careless line drops: the $2$ belongs to the whole $2 x$ , so it is raised to the fifth power as well, and the two powers of $x$ combine with opposite signs.

Illustration 3

Find the coefficient of $x^{- 2}$ in $(x - \frac{2}{x ^{2}})^{10}$ .

Write the general term and collect the power of $x$ before doing anything else.

$T_{r + 1} = (r 10) x^{10 - r} (\frac{- 2}{x ^{2}})^{r} = (r 10) (- 2)^{r} x^{10 - 3 r}$

Each step of $r$ drops the power by $3$ , which is the structural fact worth noticing.

$10 - 3 r = - 2 \Rightarrow r = 4$

$(4 10) (- 2)^{4} = 210 \times 16 = 3360$

Had the target been $x^{- 1}$ , then $10 - 3 r = - 1$ gives $r = 11/3$ , not an integer, so no such term exists and the coefficient is $0$ .

3. The Middle Term

With $n + 1$ terms, the middle depends on the parity of $n$ .

$n$	Middle term(s)
even	one term, $T_{(n /2) + 1}$
odd	two terms, $T_{(n + 1) /2}$ and $T_{(n + 3) /2}$

Illustration 4

Find the middle term or terms of $(x + 2 y)^{7}$ .

Here $n = 7$ is odd, so there are $8$ terms and two of them are middle: the $4$ th and the $5$ th.

$T_{4} = (3 7) x^{4} (2 y)^{3} = 35 \times 8 x^{4} y^{3} = 280 x^{4} y^{3}$

$T_{5} = (4 7) x^{3} (2 y)^{4} = 35 \times 16 x^{3} y^{4} = 560 x^{3} y^{4}$

The binomial coefficients are equal, $(3 7) = (4 7) = 35$ , as symmetry requires. What breaks the tie is the power of $2$ , which is why the two middle terms are not equal even though their combinations are.

4. Properties of the Binomial Coefficients

Substituting values of $x$ into $(1 + x)^{n}$ evaluates whole families of sums at once.

$\sum_{r = 0}^{n} (r n) = 2^{n}, \sum_{r = 0}^{n} (- 1)^{r} (r n) = 0$

$\sum_{r even} (r n) = \sum_{r odd} (r n) = 2^{n - 1}$

Illustration 5

For $(1 - 2 x)^{15}$ , find the sum of the coefficients and the sum of their absolute values.

Sum of the coefficients means substitute $x = 1$ into the expression itself.

$(1 - 2)^{15} = (- 1)^{15} = - 1$

The absolute values are the coefficients of $(1 + 2 x)^{15}$ , since removing the minus signs is exactly replacing $- 2 x$ by $+ 2 x$ .

$(1 + 2)^{15} = 3^{15} = 14, 348, 907$

Two sums differing by seven decimal orders, from an expression whose terms differ only in sign. That is the measure of how much cancellation the alternating signs achieve.

Illustration 6

Find $(0 15) + (2 15) + (4 15) + \dots + (14 15)$ .

Adding eight numbers, the largest of which is $(6 15) = 5005$ , is possible but slow and error-prone. Use the two substitutions instead.

$x = 1 : (0 15) + (1 15) + \dots + (15 15) = 2^{15}$

$x = - 1 : (0 15) - (1 15) + \dots - (15 15) = 0$

Adding the two equations doubles every even-indexed coefficient and annihilates every odd-indexed one.

$2 \sum_{r even} (r 15) = 2^{15} \Rightarrow \sum_{r even} (r 15) = 2^{14} = 16, 384$

Subtracting instead gives the odd-indexed sum, also $2^{14}$ . The two halves are always equal, which is another way of stating that a set has as many even-sized subsets as odd-sized ones.

Identities from calculus and from products

Differentiating $(1 + x)^{n}$ term by term and then substituting produces sums weighted by $r$ ; integrating produces sums divided by $r + 1$ . Multiplying two expansions and comparing a single coefficient produces the product identities.

$\sum_{r = 0}^{n} r (r n) = n 2^{n - 1}, \sum_{r = 0}^{n} (r n)^{2} = (n 2 n)$

The second follows from writing $(1 + x)^{n} (1 + x)^{n} = (1 + x)^{2 n}$ and comparing the coefficient of $x^{n}$ on both sides.

Illustration 7

Evaluate $(0 n) + \frac{1}{2} (1 n) + \frac{1}{3} (2 n) + \dots + \frac{1}{n + 1} (n n)$ .

A denominator of $r + 1$ against a coefficient $(r n)$ is the signature of an integration, because integrating $x^{r}$ raises the power to $r + 1$ and divides by it.

$\int_{0}^{1} (1 + x)^{n} d x = \int_{0}^{1} \sum_{r = 0}^{n} (r n) x^{r} d x = \sum_{r = 0}^{n} \frac{1}{r + 1} (r n)$

The left side is elementary.

$\int_{0}^{1} (1 + x)^{n} d x = [\frac{( 1 + x ) ^{n + 1}}{n + 1}]_{0}^{1} = \frac{2 ^{n + 1} - 1}{n + 1}$

$\sum_{r = 0}^{n} \frac{1}{r + 1} (r n) = \frac{2 ^{n + 1} - 1}{n + 1}$

Check it at $n = 2$ : the sum is $1 + \frac{1}{2} (2) + \frac{1}{3} (1) = \frac{7}{3}$ , and the formula gives $(8 - 1) /3 = \frac{7}{3}$ .

The general rule is worth stating. A factor of $r$ in front of $(r n)$ points to differentiation; a divisor of $r + 1$ points to integration; a squared coefficient points to comparing coefficients in a product.

5. Greatest Coefficient and Greatest Term

These are different questions, and confusing them is the standard error here.

Question	Depends on
Greatest coefficient	$n$ only
Greatest term	$n$ and the value of $x$

The greatest coefficient of $(1 + x)^{n}$ is the middle one, because the coefficients rise to the centre and fall away symmetrically. The greatest term can sit anywhere, since a large power of $x$ can outweigh a small combination.

Both are found the same way, from the ratio of consecutive terms.

$\frac{T _{r + 2}}{T _{r + 1}} = \frac{n - r}{r + 1} \cdot x$

Terms increase while that ratio exceeds $1$ and decrease after, so the greatest term is at the change of direction.

Illustration 8

Show, from the ratio, that the greatest coefficient of $(1 + x)^{n}$ is the middle one.

Set $x = 1$ so only the combinations remain.

$\frac{( r + 1 n )}{( r n )} = \frac{n - r}{r + 1}$

$\frac{n - r}{r + 1} > 1 ⟺ n - r > r + 1 ⟺ r < \frac{n - 1}{2}$

So the coefficients strictly increase while $r$ is below $(n - 1) /2$ and strictly decrease after, with the turning point at the centre.

For even $n$ that gives one peak, at $r = n /2$ ; for odd $n$ the ratio equals $1$ exactly once, producing two equal greatest coefficients side by side.

Notice that this argument never computed a single factorial. The ratio of consecutive terms is nearly always easier than the terms themselves.

Illustration 9

Find the greatest term in the expansion of $(2 + 3 x)^{9}$ when $x = \frac{1}{3}$ .

Never compute ten terms. Form the ratio and ask where it drops below one.

$\frac{T _{r + 2}}{T _{r + 1}} = \frac{9 - r}{r + 1} \cdot \frac{3 x}{2} = \frac{9 - r}{r + 1} \cdot \frac{1}{2}$

$\frac{9 - r}{2 ( r + 1 )} > 1 ⟺ 9 - r > 2 r + 2 ⟺ r < \frac{7}{3}$

So the terms increase for $r = 0, 1, 2$ and decrease from $r = 3$ onwards. The last increase produces $T_{4}$ .

$T_{4} = (3 9) 2^{6} (3 \times \frac{1}{3})^{3} = 84 \times 64 \times 1 = 5376$

The greatest coefficient of this expansion would have been at the middle, near $r = 4$ , so the peak has moved left by one place. The powers of $2$ falling faster than the powers of $3 x$ rise is what shifted it.

6. Applications: Divisibility and Remainders

Split the base so that one piece is the divisor. Every term containing that piece is then divisible, and only the last term survives.

$(k + 1)^{n} = all divisible by k (0 n) k^{n} + \dots + (n - 1 n) k + 1$

Illustration 10

Show that $2^{3 n} - 1$ is divisible by $7$ for every positive integer $n$ .

Rewrite the power so that $7$ appears in the base.

$2^{3 n} = (2^{3})^{n} = 8^{n} = (7 + 1)^{n}$

$(7 + 1)^{n} = (0 n) 7^{n} + (1 n) 7^{n - 1} + \dots + (n - 1 n) 7 + 1$

Every term except the last carries a factor of $7$ , so the whole sum is $7 m + 1$ for some integer $m$ .

$2^{3 n} - 1 = 7 m$

The technique is entirely in the first line. Recognising $2^{3 n}$ as $8^{n}$ , and $8$ as $7 + 1$ , is what makes the expansion collapse.

Last digits

Last digits are remainders modulo $10$ , $100$ and so on, so the same split applies with a base built from a power of ten.

Illustration 11

Find the last two digits of $7^{50}$ .

Last two digits means the remainder modulo $100$ , so look for a power of $7$ close to a multiple of $100$ . Since $7^{2} = 49$ , the base $50 - 1$ is the one that helps.

$7^{50} = (7^{2})^{25} = 4 9^{25} = (50 - 1)^{25}$

Expand and note that any term carrying $5 0^{2} = 2500$ is divisible by $100$ and therefore contributes nothing.

$(50 - 1)^{25} \equiv (1 25) 50 (- 1)^{24} + (- 1)^{25} (mod 100)$

$\equiv 1250 - 1 \equiv 50 - 1 = 49 (mod 100)$

The last two digits are $49$ . Only two terms of a twenty-six term expansion survived, and choosing the base $50 - 1$ rather than $49$ is what killed the rest.

Illustration 12

Find the remainder when $5^{99}$ is divided by $13$ .

Look for a power of $5$ that lands near a multiple of $13$ . Since $5^{2} = 25$ and $26 = 2 \times 13$ , the base $25 = 26 - 1$ is the useful one.

$5^{99} = 5 \times (5^{2})^{49} = 5 (26 - 1)^{49}$

Expanding $(26 - 1)^{49}$ , every term but the last carries a factor of $26$ , and the last is $(- 1)^{49} = - 1$ .

$(26 - 1)^{49} = 26 k - 1 \Rightarrow 5^{99} = 130 k - 5$

$130 k is divisible by 13, so 5^{99} \equiv - 5 \equiv 8 (mod 13)$

The remainder is $8$ . Note the last step: a negative remainder is corrected by adding the divisor once, and $- 5 + 13 = 8$ .

7. Sums, Differences and Rational Terms

Adding or subtracting two expansions

For a surd $d$ , expanding $(p + d)^{n}$ and $(p - d)^{n}$ gives terms that differ only in sign wherever the power of $d$ is odd.

$(p + d)^{n} + (p - d)^{n} = 2 [terms with even powers of d] = an integer$

$(p + d)^{n} - (p - d)^{n} = 2 d [an integer]$

The difference keeps only the odd powers, so it is an integer multiple of $d$ rather than an integer. Which of the two you need is decided by whether the question asks for an integer or for a surd.

Illustration 13

Show that $(2 + 3)^{5}$ is just below an integer, and find its integer part.

Its conjugate $(2 - 3)$ is about $0.268$ , so the fifth power is tiny, roughly $0.0014$ . Add the two and the surds cancel.

$(2 + 3)^{5} + (2 - 3)^{5} = 2 [(0 5) 2^{5} + (2 5) 2^{3} (3) + (4 5) 2 (9)]$

$= 2 [32 + 240 + 90] = 2 (362) = 724$

Since the conjugate term is positive but under $0.002$ :

$(2 + 3)^{5} = 724 - 0.0014 \dots \Rightarrow integer part 723$

The fractional part is $0.9986$ , which is the real content of the question: the conjugate power measures exactly how close to the integer you are.

Counting rational terms

In an expansion of $(a^{1/ p} + b^{1/ q})^{n}$ , a term is rational only when both fractional powers clear simultaneously. Write the general term, impose one divisibility condition on $r$ from each surd, and count the $r$ in range satisfying both.

Illustration 14

How many rational terms are there in the expansion of $(2^{1/3} + 3^{1/2})^{12}$ ?

Write the general term and read off both fractional powers.

$T_{r + 1} = (r 12) (2^{1/3})^{12 - r} (3^{1/2})^{r} = (r 12) 2^{(12 - r) /3} 3^{r /2}$

Rationality demands both exponents be whole numbers, and the two conditions must hold at the same time.

$3 ∣ (12 - r) \Rightarrow r \equiv 0 (mod 3), 2 ∣ r \Rightarrow r even$

Both together mean $r$ is a multiple of $6$ , and $r$ runs from $0$ to $12$ .

$r = 0, 6, 12 \Rightarrow 3 rational terms$

Satisfying one condition is not enough, and that is where marks go: $r = 3$ clears the cube root but leaves $3^{3/2}$ behind.

8. Approximation

When $x$ is small, terms with high powers of $x$ become negligible and the expansion is truncated.

$(1 + x)^{n} \approx 1 + n x for ∣ x ∣ small$

Illustration 15

Estimate $(1.02)^{10}$ with one correction term and then with two, and compare with the true value $1.21899$ .

Write $1.02 = 1 + 0.02$ so that $x$ is small.

$(1 + 0.02)^{10} \approx 1 + 10 (0.02) = 1.20$

That is already within $2$ per cent. Adding the next term sharpens it considerably.

$+ (2 10) (0.02)^{2} = 45 \times 0.0004 = 0.018 \Rightarrow 1.218$

$error: 1.21899 - 1.218 = 0.00099$

The third term would contribute $(3 10) (0.02)^{3} = 120 \times 0.000008 = 0.00096$ , recovering almost all of what is left. Each successive term is roughly fifty times smaller, which is why truncating early is safe here and would not be if $x$ were near $1$ .

Summary

The binomial theorem is a counting statement: the coefficient of $a^{n - r} b^{r}$ counts which $r$ of the $n$ brackets supplied the $b$ .

There are $n + 1$ terms; powers of $a$ fall, powers of $b$ rise, the two always sum to $n$ , and the coefficients are symmetric, which is a free error check.

$T_{r + 1} = (r n) a^{n - r} b^{r}$ , and the first term has $r = 0$ . The $4$ th term uses $r = 3$ .

For any "find the coefficient of" question: write the general term, collect the power of $x$ , set it equal to the target. A non-integral $r$ means the coefficient is zero.

Even $n$ has one middle term, odd $n$ has two, and the two are unequal whenever the binomial carries constants.

Sum of the coefficients means substituting $x = 1$ into the expression, so $(2 + 3 x)^{20}$ sums to $5^{20}$ and not $2^{20}$ .

$x = - 1$ gives zero for $(1 + x)^{n}$ ; averaging the two substitutions splits the coefficients into even and odd halves, each $2^{n - 1}$ .

Differentiating gives sums weighted by $r$ , and comparing coefficients in a product of two expansions gives the squared-coefficient identities.

Greatest coefficient depends on $n$ alone and sits at the middle; greatest term depends on $x$ as well and can sit anywhere. Both come from the ratio $(n - r) x / (r + 1)$ .

A factor of $r$ in front of a coefficient points to differentiation, a divisor of $r + 1$ points to integration, and a squared coefficient points to a product of two expansions.

A term is rational only when every fractional exponent clears at once, so impose one divisibility condition per surd and intersect them.

For divisibility, split the base so the divisor appears in it: $2^{3 n}$ becomes $(7 + 1)^{n}$ , and every term but the last is divisible.

Adding a surd expansion to its conjugate cancels the odd powers and leaves an integer, which locates the fractional part exactly.

For approximation, $(1 + x)^{n} \approx 1 + n x$ when $x$ is small, and each further term shrinks by roughly a factor of $x$ times $n$ .

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

The organising principle

t h ecoe f f i c i e n t o f a^{n - r} b^{r} co u n t s w hi c h r o f t h e nb r a c k e t ss u ppl i e d t h e b

That is why the coefficients are combinations, why they are symmetric, and why substituting a value of x evaluates a whole family of sums at once.

The theorem

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

There are n+1 terms; powers of a fall while powers of b rise and the two always total n. The coefficients are palindromic, which is a free error check on any expansion.

General term

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

The first term has r = 0, so the 4th term of (a+b)^10 uses r = 3 and is 120 a^7 b^3, not 210 a^6 b^4. This off-by-one is the single most common slip in the chapter.

Finding a specified coefficient

w r i t e T_{r + 1}, co l l ec tt h e p o w er o f x, se t i t e q u a l t o t h e t a r g e t an d so l v e f or r

A non-integral r, or one outside 0 to n, means that power does not occur and the coefficient is zero. That is a legitimate answer rather than an error.

Middle term

n e v e n : o n e mi dd l e t er m T_{n /2 + 1}; n o dd : tw o, T_{(n + 1) /2} an d T_{(n + 3) /2}

For (x+2y)^7 the two middle terms have equal combinations, 35 each, but different values, 280 and 560, because the powers of 2 break the tie.

Sums by substitution

x = 1 gives the sum of all coefficients; x = -1 gives the alternating sum, which is 0 for (1+x)^n

Substitute into the expression. For (2+3x)^20 the sum is 5^20 and for (1-2x)^15 it is -1, while the sum of absolute values is 3^15.

Even and odd splits

\sum_{r even} (r n) = \sum_{r odd} (r n) = 2^{n - 1}

Add the x = 1 and x = -1 equations to isolate the even-indexed coefficients, subtract them to isolate the odd. A set has as many even-sized subsets as odd-sized ones.

Identities from calculus

\sum r (r n) = n 2^{n - 1}; \sum \frac{1}{r + 1} (r n) = \frac{2 ^{n + 1} - 1}{n + 1}

A factor of r in front points to differentiation, a divisor of r+1 points to integration, and a squared coefficient points to comparing coefficients in a product.

Product identity

\sum_{r} (r n)^{2} = (n 2 n)

Comes from writing (1+x)^n (1+x)^n = (1+x)^{2n} and comparing the coefficient of x^n on both sides. The same trick gives the general Vandermonde identity.

Ratio of consecutive terms

T_{r + 2} / T_{r + 1} = \frac{( n - r )}{( r + 1 )} x

Terms rise while the ratio exceeds 1 and fall after. This locates both the greatest coefficient, which depends on n alone and sits at the middle, and the greatest term, which depends on x too and can sit anywhere.

Divisibility and remainders

split the base so the divisor appears in it, then every term but the last carries it

2^{3n} becomes (7+1)^n, proving divisibility by 7. For 5^{99} modulo 13, write it as 5 times (26-1)^{49}. Correct a negative remainder by adding the divisor once.

Surd conjugate pairs

(p + d)^{n} + (p - d)^{n} i s anin t e g er; t h e d i f f er e n ce i s anin t e g er m u l t i pl eo f d

Adding cancels the odd powers of the surd. Since (2 - root3)^5 is under 0.002, the sum 724 places (2 + root3)^5 just below 724 with integer part 723.

Approximation for small x

(1 + x)^{n} \approx 1 + n x, an d t h e n e x tt er mi s (2 n) x^{2}

For (1.02)^{10} one term gives 1.20 and two give 1.218 against a true 1.21899. Each successive term shrinks by roughly nx, which is why truncating early is safe only when x is small.

JEE Main test series

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Binomial Theorem shows up in every JEE Main paper — sit a timed one and find out where you actually stand.

~1Est. Qs

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Traps JEE Main sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Answering the sum of the coefficients as 2 to the n whatever the binomial is

✓ Substitute x equal to 1 into the actual expression. For (2+3x)^20 that gives 5^20, not 2^20, because the coefficient of x^r is C(20,r) times 2^{20-r} times 3^r and the powers carry nearly all the size.

Why it happens: The result is learnt from (1+x)^n, where the combinations really are the coefficients.

WATCH OUT

✗ Using r equal to the term number in the general term

✓ The first term has r = 0, so the kth term uses r = k - 1. The 4th term of (a+b)^10 is C(10,3) a^7 b^3 = 120 a^7 b^3, while C(10,4) gives a different term and a different number, 210 a^6 b^4.

Why it happens: The formula is written with T_{r+1} but read as though the two indices agreed.

WATCH OUT

✗ Forgetting to raise the whole of a compound term to its power

✓ In (2x + 1/x)^9 the fifth term is C(9,4) times (2x)^5 times x^{-4}, and the 2 must be raised to the fifth power too, giving 4032x. Write the bracket out with its coefficient inside before expanding anything.

Why it happens: The letter is what the eye tracks, and the coefficient in front of it is easy to leave behind.

WATCH OUT

✗ Confusing the greatest coefficient with the greatest term

✓ The greatest coefficient depends on n alone and sits at the middle. The greatest term depends on x as well, since a large power of x can outweigh a smaller combination. Form the ratio (n-r)x/(r+1) and find where it falls below one.

Why it happens: For (1+x)^n at x = 1 the two coincide, and that is the case most often met.

WATCH OUT

✗ Declaring a term rational after clearing only one of the surds

✓ Every fractional exponent must clear at the same value of r. In the expansion of (2^{1/3} + 3^{1/2})^{12} the conditions are r divisible by 3 and r even, so r must be a multiple of 6, and there are three rational terms, not six.

Why it happens: Each divisibility condition is checked as it appears, and the first success feels like the answer.

WATCH OUT

✗ Choosing a base that is not adjacent to a multiple of the divisor

✓ The expansion collapses only when the base is the divisor plus or minus one. For the last two digits of 7^{50}, write it as (50-1)^{25} rather than 49^{25}: every term with 50 squared is divisible by 100 and only two terms survive.

Why it happens: Any rewriting of the power feels equally good, so the first one that comes to mind is used.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Binomial Theorem?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•The coefficient is a count of which brackets supplied the second letter
•There are n+1 terms and the coefficients are palindromic, which checks any expansion
•T_{r+1} = C(n,r) a^{n-r} b^r, and the first term has r = 0
•For a specified coefficient: write the general term, collect the power, solve for r
•A non-integral r means the coefficient is zero, and that is an answer
•Even n has one middle term, odd n has two and they are unequal when constants are present
•Sum of coefficients means substituting x = 1 into the expression, not into (1+x)^n
•x = 1 and x = -1 together split the coefficients into two halves of 2^{n-1} each
•A factor of r means differentiate; a divisor of r+1 means integrate; a squared coefficient means a product
•Greatest coefficient depends on n; greatest term depends on n and x; both come from the ratio
•For divisibility, rewrite the base as divisor plus or minus one and only the last term survives
•Adding a surd expansion to its conjugate gives an integer and pins down the fractional part

JEE Main question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: 4

Question style	Marks each	Typical count
General term and specified coefficients	2	1
Coefficient identities and sums	1	1
Divisibility, remainders and greatest term	1	1

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Write the general term before doing anything else, then collect the power of x and set it equal to the target. Almost every question in this chapter is that one routine with a different target.
For any sum of coefficients, decide which substitution you need and put it into the expression exactly as printed. Never substitute into a remembered (1+x)^n.
Count the terms as n+1 and check the coefficients are palindromic. Both are one-second checks and both catch the most common expansion errors.
Use the ratio of consecutive terms whenever a question asks for a greatest anything. Computing the terms themselves is slower and rarely necessary.
For divisibility, remainder and last-digit questions, spend the first line rewriting the base as divisor plus or minus one. If the rewriting does not produce that shape, you have chosen the wrong split.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

The binomial probability distribution is this expansion w…

The binomial probability distribution is this expansion with a and b replaced by the probabilities of success and failure, so the coefficients count the orderings in which a given number of successes can occur

Cryptography and computer arithmetic use the divisibility…

Cryptography and computer arithmetic use the divisibility split constantly, since fast modular exponentiation is exactly the observation that only the last term of an expansion survives modulo the chosen base

Engineering approximations replace awkward powers by thei…

Engineering approximations replace awkward powers by their first two binomial terms, which is how small corrections for thermal expansion, relativistic speed or interest compounding are estimated in one line

Where else this topic is tested

Prepare once, score in every exam that asks it.

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Because the expansion is an identity, true for every value of x, so both sides may be evaluated anywhere. Setting x equal to 1 makes every power of x equal to 1, so the right side collapses to the bare sum of the coefficients while the left side is a single number you can compute. The same reasoning at x equal to minus 1 inserts alternating signs, and averaging the two results isolates the even-indexed coefficients. The essential discipline is to substitute into the expression as given. For (2+3x)^20 the left side becomes 5^20, and anyone who substitutes into a remembered (1+x)^n instead gets 2^20 and is wrong by seven orders of magnitude.

Look at what is attached to the binomial coefficient. A factor of r in front, as in the sum of r times C(n,r), means differentiation, because differentiating x^r brings down an r. A divisor of r+1, as in the sum of C(n,r) over r+1, means integration, because integrating x^r divides by r+1. A squared coefficient, or a product of two coefficients from different rows, means writing a product of two expansions and comparing the coefficient of one power on both sides, which is how the sum of squared coefficients turns out to be C(2n,n). Signs alternating suggests substituting minus x somewhere in the process. Reading the shape first saves you from trying every technique in turn.

Because the middle term has the largest combination, but the term is the combination multiplied by powers of both parts of the binomial, and those powers change from term to term. If x is small, later terms carry high powers of a small number and shrink faster than the combinations grow, so the peak moves left. If x is large the peak moves right. Only when the two parts are balanced, as in (1+x)^n at x equal to 1, does the peak sit at the middle. The reliable method is the ratio of consecutive terms, (n-r)x over (r+1), which is greater than 1 exactly while the terms are still rising, so the last r satisfying that inequality gives the greatest term.

Rewrite the power so that the base is one more or one less than the divisor, or than a convenient multiple of it. Once the base is that shape, expanding leaves every term carrying a factor of the divisor except the final one, which is plus or minus 1. So 2 to the 3n becomes 8 to the n, then (7+1) to the n, which is 7 times an integer plus 1, proving divisibility of 2^{3n} minus 1 by 7. For 5 to the 99 modulo 13, notice 5 squared is 25 and 26 is twice 13, so write the whole thing as 5 times (26-1) to the 49. Two details catch people: pull out any leftover factor first, and if the surviving remainder is negative, add the divisor once to bring it into range.

The syllabus states the theorem for a positive integral index, and that is what the questions are built on. A positive whole-number index is exactly what makes the expansion finite, with n+1 terms, and what makes every coefficient a genuine count of which brackets supplied which letter. The approximation work in this chapter stays inside that scope: (1.02) to the tenth is a positive integral index applied to a base close to 1, and truncating after two or three terms is a matter of the later terms being small, not of an infinite series. If you meet an infinite expansion in other material, treat it as background rather than as examinable technique here.