Chapter 5 of 16

31%

JEE Advanced Mathematics★ ~1 question across Paper 1 and Paper 2 combined, usually a coefficient extraction from a product or a binomial sum proved by comparing coefficients

Binomial Theorem

The general term as the only tool, greatest term against greatest coefficient, coefficient extraction, series with a fractional index and where they stop being true.

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JEE Advanced PYQsSolved previous-year questions

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

1Write the general term of any binomial expansion and use it to locate a specified term, a term independent of the variable, or a rational term
2Distinguish the greatest coefficient, which depends only on the index, from the greatest term, which depends on the value of the variable
3Extract a coefficient from a product or a geometric sum of expansions without expanding each one
4Prove binomial identities by comparing coefficients on both sides of a product, including Vandermonde's identity
5Sum binomial series by substitution, by differentiation or integration, by the identity $r (r n) = n (r - 1 n - 1)$ , and by reversing the sum
6State the range of validity of an expansion with a negative or fractional index, and use expansions to find remainders and prove divisibility

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Why this chapter matters in JEE Advanced

The binomial theorem is one of the few chapters where a single mechanical device, the general term, answers almost every question, so the marks go to candidates who know what to do with it rather than to those who recall formulas. Advanced adds two things Main leaves out: expansions for a negative or fractional index, which carry a validity condition that changes the answer completely if ignored, and coefficient identities that are proved by comparing two expansions rather than by summing anything. The chapter also underpins the series work in Sequences and the approximations used throughout Physics.

Before you start — revise these

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Combinations and the meaning of

(r n)

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Geometric series and their sums

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Differentiation and integration of simple polynomials

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Modular arithmetic at the level of remainders on division

Binomial Theorem

Expanding $(1 + x)^{- 1}$ gives $1 - x + x^{2} - x^{3} + \dots$ . Put $x = 2$ and read off the sum.

The left side is $(1 + 2)^{- 1} = \frac{1}{3}$ . The right side is $1 - 2 + 4 - 8 + \dots$ , which has no sum at all.

The expansion for a negative or fractional index is not an identity between two polynomials. It is an infinite series, and it represents the function only when $∣ x ∣ < 1$ . Outside that range the series diverges and the equation is meaningless, however comfortable the algebra looks.

For a positive integer index no such condition exists, because the expansion terminates after $n + 1$ terms and is a genuine polynomial identity. That is the fault line running through this chapter: the same-looking formula behaves completely differently depending on the index, and Advanced tests the difference directly by asking for the range of validity or by choosing a value of $x$ near the boundary.

1. Everything from the general term

For a positive integer $n$ ,

$(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}, T_{r + 1} = (r n) a^{n - r} b^{r} .$

The subscript convention matters: $T_{r + 1}$ carries $b$ to the power $r$ , so the sixth term has $r = 5$ . Nearly every question in this chapter is answered by writing the general term, simplifying the power of $x$ to a single exponent, and setting that exponent to whatever the question wants.

Two structural facts about the coefficients come from a single combinatorial reading. Choosing $r$ objects from $n + 1$ either uses a fixed special object or does not, giving Pascal's rule

$(r - 1 n) + (r n) = (r n + 1),$

and choosing $r$ to keep is the same as choosing $n - r$ to discard, giving the symmetry $(r n) = (n - r n)$ .

The theorem itself is proved from that rule by induction: assuming the expansion for $n$ , multiplying by one more factor of $(1 + x)$ adds each coefficient to its neighbour, which is precisely what Pascal's rule says the next row does.

Illustration 1

Find the term independent of $x$ in $(x^{2} - \frac{1}{x})^{12}$ .

$T_{r + 1} = (r 12) (x^{2})^{12 - r} (- \frac{1}{x})^{r} = (- 1)^{r} (r 12) x^{24 - 3 r} .$

Setting $24 - 3 r = 0$ gives $r = 8$ , so the term is $(8 12) = 495$ . The sign is positive because $r$ is even, and checking the parity of $r$ is worth the second it takes.

Illustration 2

Find the middle term of $(2 x - \frac{3}{x})^{10}$ .

With $n = 10$ there is one middle term, the sixth, at $r = 5$ :

$(5 10) (2 x)^{5} (- \frac{3}{x})^{5} = 252 \times 32 \times (- 243) = - 1959552,$

and the powers of $x$ cancel, as they must for a symmetric pair of exponents. When $n$ is odd there are two middle terms, at $r = \frac{n - 1}{2}$ and $r = \frac{n + 1}{2}$ .

2. Greatest coefficient and greatest term are different questions

The greatest coefficient is a property of $n$ alone: the binomial coefficients rise to the middle and fall symmetrically, so the largest is $(n /2 n)$ for even $n$ , and the equal pair $(( n \pm 1 ) /2 n)$ for odd $n$ .

The greatest term depends on $x$ as well, because the powers of $x$ can outweigh the coefficients. Compare consecutive terms:

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

which exceeds $1$ while $r < \frac{( n + 1 ) x}{1 + x}$ . The terms therefore increase and then decrease, and the turning point is found by taking the integer part of that bound. When the bound is exactly an integer, two consecutive terms are equal and both are greatest.

The reason the coefficients rise and then fall is the same ratio, taken at $x = 1$ : consecutive coefficients are in the ratio $\frac{n - r + 1}{r}$ , which exceeds $1$ exactly while $r < \frac{n + 1}{2}$ . So there is a single peak, at the middle, and no coefficient is ever a local dip.

For an expansion such as $(a + b x)^{n}$ the phrase "greatest coefficient" is ambiguous and questions say which they mean: the numerically largest coefficient folds the constants $a$ and $b$ into the ratio, and is found by the same comparison with $x$ replaced by $b / a$ .

Illustration 3

In $(1 + x)^{15}$ , find the greatest coefficient and, separately, the greatest term when $x = \frac{1}{3}$ .

Since $15$ is odd, the greatest coefficients are $(7 15) = (8 15) = 6435$ , sitting at the eighth and ninth terms.

For the greatest term, $\frac{( n + 1 ) x}{1 + x} = \frac{16 \cdot \frac{1}{3}}{\frac{4}{3}} = 4$ . The bound is exactly $4$ , so $T_{5} = T_{4}$ and both are greatest — at the fourth and fifth terms, nowhere near the middle. Reporting the middle term as the greatest is the standard error, and it is only correct when $x = 1$ .

3. Extracting a coefficient from a product

A coefficient of $x^{k}$ in a product of expansions is a sum over the ways the exponents can add to $k$ . Two devices make almost every such question short: sum a geometric series of binomials before expanding, and convert negative powers of $x$ into a single shifted expansion.

Illustration 4

Find the coefficient of $x^{5}$ in $(1 + x)^{21} + (1 + x)^{22} + \dots + (1 + x)^{30}$ .

The sum is geometric with ratio $(1 + x)$ :

$\frac{( 1 + x ) ^{31} - ( 1 + x ) ^{21}}{( 1 + x ) - 1} = \frac{( 1 + x ) ^{31} - ( 1 + x ) ^{21}}{x} .$

So the coefficient of $x^{5}$ here is the coefficient of $x^{6}$ in the numerator, namely $(6 31) - (6 21)$ . Summing ten expansions term by term would take ten times as long and offer ten chances to slip.

Illustration 5

Find the coefficient of $x^{r}$ in $(1 + x)^{n} (1 + \frac{1}{x})^{n}$ .

Write the second factor as $\frac{( 1 + x ) ^{n}}{x ^{n}}$ , so the product is $\frac{( 1 + x ) ^{2 n}}{x ^{n}}$ . The coefficient of $x^{r}$ is therefore the coefficient of $x^{n + r}$ in $(1 + x)^{2 n}$ , which is $(n + r 2 n)$ .

Illustration 6

Prove that $r = 0 \sum n (r n)^{2} = (n 2 n)$ .

Compare coefficients of $x^{n}$ on both sides of $(1 + x)^{n} (1 + x)^{n} = (1 + x)^{2 n}$ . On the left, a term $x^{n}$ arises from $x^{r}$ in the first factor and $x^{n - r}$ in the second, contributing $(r n) (n - r n) = (r n)^{2}$ by symmetry. Summing over $r$ gives the result.

The same argument with unequal indices gives Vandermonde's identity $\sum_{r} (r m) (k - r n) = (k m + n)$ , which reads as choosing $k$ people from two rooms by first deciding how many come from each.

4. Summing a binomial series

Three techniques cover the standard sums, and choosing between them is decided by what multiplies the coefficient.

If the coefficient stands alone, substitute a value of $x$ into $(1 + x)^{n}$ . Putting $x = 1$ gives $\sum (r n) = 2^{n}$ and $x = - 1$ gives the alternating sum $0$ .

If the coefficient is multiplied by $r$ , differentiate first. From $\sum (r n) x^{r} = (1 + x)^{n}$ ,

$\sum r (r n) x^{r - 1} = n (1 + x)^{n - 1} \Rightarrow \sum r (r n) = n 2^{n - 1} .$

The same result follows from the identity $r (r n) = n (r - 1 n - 1)$ , which says that choosing a committee of $r$ with a chair is the same as choosing the chair first.

If the coefficient is divided by $r + 1$ , integrate instead. Integrating from $0$ to $1$ ,

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

A fourth technique costs nothing and often replaces all three. Because $(r n) = (n - r n)$ , a sum can be written forwards and backwards and the two versions added. Taking $S = \sum r (r n)$ and rewriting it with $r$ replaced by $n - r$ gives $S = \sum (n - r) (r n)$ , so

$2 S = \sum n (r n) = n 2^{n} \Rightarrow S = n 2^{n - 1},$

recovering the earlier result without calculus. Reversing the sum is the first thing to try whenever the multiplier is linear in $r$ , and it doubles as an independent check on an answer obtained by differentiating.

Illustration 7

Evaluate $r = 0 \sum n r^{2} (r n)$ .

Write $r^{2} = r (r - 1) + r$ . The first part gives $n (n - 1) \sum (r - 2 n - 2) = n (n - 1) 2^{n - 2}$ using the identity twice, and the second gives $n 2^{n - 1}$ . Adding,

$n (n - 1) 2^{n - 2} + n 2^{n - 1} = n 2^{n - 2} (n - 1 + 2) = n (n + 1) 2^{n - 2} .$

Testing at $n = 2$ : the sum is $0 + 1 \cdot 2 + 4 \cdot 1 = 6$ , and the formula gives $2 \cdot 3 \cdot 1 = 6$ .

5. More than two terms

The multinomial expansion of $(x_{1} + x_{2} + \dots + x_{k})^{n}$ has general term

$\frac{n !}{n _{1} ! n _{2} ! \dots n _{k} !} x_{1}^{n_{1}} \dots x_{k}^{n_{k}}, \sum n_{i} = n,$

and the number of distinct terms is the number of ways to write $n$ as an ordered sum of $k$ non-negative parts, namely $(k - 1 n + k - 1)$ by stars and bars.

Illustration 8

How many terms does $(x + y + z)^{10}$ have, and what is the coefficient of $x^{2} y^{3} z^{5}$ ?

The number of terms is $(2 12) = 66$ . The coefficient is $\frac{10 !}{2 ! 3 ! 5 !} = 2520$ .

For a trinomial raised to a small power it is often quicker to group two terms and apply the ordinary binomial theorem twice, but the multinomial coefficient avoids the bookkeeping entirely.

6. Negative and fractional indices

For any real $n$ and $∣ x ∣ < 1$ ,

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

The numerators are falling products rather than factorials, so nothing cancels to zero unless $n$ is a non-negative integer — which is exactly why the series terminates in that case and not otherwise. Two special cases are worth carrying: $(1 - x)^{- 1} = 1 + x + x^{2} + \dots$ and $(1 - x)^{- 2} = 1 + 2 x + 3 x^{2} + \dots$ , the second being the derivative of the first.

The general term for a negative integer index is worth stating separately, because it appears constantly in counting arguments. Expanding $(1 - x)^{- n}$ gives

$(1 - x)^{- n} = \sum_{r \geq 0} (r n + r - 1) x^{r},$

whose coefficients are exactly the stars-and-bars counts: the number of ways to write $r$ as an ordered sum of $n$ non-negative parts. The connection is not a coincidence — multiplying $n$ copies of $1 + x + x^{2} + \dots$ chooses one power from each factor, which is the same decision as filling $n$ boxes.

Illustration 9

Estimate $1.02$ to five decimal places.

Take $n = \frac{1}{2}$ and $x = 0.02$ , comfortably inside the range of validity:

$1 + \frac{1}{2} (0.02) - \frac{1}{8} (0.02)^{2} = 1 + 0.01 - 0.00005 = 1.00995.$

The next term contributes about $1 0^{- 6}$ , so the estimate is good to five places. The true value is $1.0099504 \dots$

Illustration 10

For what values of $x$ is the expansion of $(2 + 3 x)^{- 2}$ valid, and what is its third term?

Factor out the constant first: $(2 + 3 x)^{- 2} = \frac{1}{4} (1 + \frac{3 x}{2})^{- 2}$ , which requires $\frac{3 x}{2} < 1$ , that is $∣ x ∣ < \frac{2}{3}$ .

The third term of $(1 + u)^{- 2}$ is $\frac{( - 2 ) ( - 3 )}{2 !} u^{2} = 3 u^{2}$ , so here it is $\frac{1}{4} \cdot 3 (\frac{3 x}{2})^{2} = \frac{27 x ^{2}}{16}$ .

Factoring to make the leading term $1$ is not cosmetic — the validity condition can only be read off once the expression is in that form.

7. Divisibility and remainders

Writing a large power as $(multiple of m \pm 1)^{k}$ and expanding makes every term except the last divisible by $m$ . This turns most remainder questions into one line.

Illustration 11

Find the remainder when $2^{50}$ is divided by $7$ .

$2^{50} = 4 \cdot 2^{48} = 4 \cdot (2^{3})^{16} = 4 (7 + 1)^{16} .$

Every term of $(7 + 1)^{16}$ except the last carries a factor of $7$ , so $(7 + 1)^{16} \equiv 1$ and the remainder is $4$ .

The same method works with a minus sign, and then the parity of the exponent decides the answer. Expanding $5^{99} = (6 - 1)^{99}$ , every term except the last carries a factor of $6$ , and the last is $(- 1)^{99} = - 1$ . So $5^{99} \equiv - 1 \equiv 5$ modulo $6$ . Had the exponent been even, the remainder would have been $1$ instead, which is why the parity has to be read off before the expansion is discarded.

Illustration 12

Find the last two digits of $3^{400}$ .

$3^{400} = (3^{4})^{100} = 8 1^{100} = (80 + 1)^{100} .$

Modulo $100$ , every term with $8 0^{2}$ or higher vanishes, leaving $1 + 100 \times 80$ , and $100 \times 80$ is itself a multiple of $100$ . So $3^{400} \equiv 1$ and the last two digits are $01$ .

Illustration 13

Show that $3^{2 n + 2} - 8 n - 9$ is divisible by $64$ for every positive integer $n$ .

Write $3^{2 n + 2} = 9^{n + 1} = (1 + 8)^{n + 1}$ and expand:

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

Subtracting $8 n + 9$ removes exactly the first two terms, and every remaining term carries $8^{2} = 64$ as a factor. The choice to expand about $1 + 8$ rather than $9$ is what makes the two unwanted terms cancel.

Illustration 14

Find the coefficient of $x^{3}$ in the expansion of $\frac{1 + 2 x}{( 1 - x ) ^{2}}$ , stating where it is valid.

Using $(1 - x)^{- 2} = \sum_{k \geq 0} (k + 1) x^{k}$ , valid for $∣ x ∣ < 1$ , the coefficient of $x^{3}$ in the product is the contribution from $1 \times 4 x^{3}$ and from $2 x \times 3 x^{2}$ , giving $4 + 6 = 10$ .

8. Reading a polynomial at $x = 1$ and $x = - 1$

Substituting a single value into a whole expansion answers questions that would otherwise need every coefficient. The sum of all coefficients of a polynomial $f (x)$ is $f (1)$ , because setting $x = 1$ leaves each coefficient contributing once. Setting $x = - 1$ instead attaches a minus sign to every odd power, so

$even-power coefficients = \frac{f ( 1 ) + f ( - 1 )}{2}, odd-power coefficients = \frac{f ( 1 ) - f ( - 1 )}{2} .$

Applied to $(1 + x)^{n}$ these give $2^{n}$ and $0$ , so the even-indexed and odd-indexed binomial coefficients each sum to $2^{n - 1}$ . The same pair of substitutions explains why $(1 + x)^{n} + (1 - x)^{n}$ keeps only the even powers, and so has $\frac{n}{2} + 1$ terms when $n$ is even and $\frac{n + 1}{2}$ when $n$ is odd.

Ratios of adjacent coefficients are the other standard handle. Since

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

any question that supplies the ratio of consecutive coefficients supplies a linear equation in $n$ and $r$ , and two such ratios determine both.

Illustration 15

Three consecutive coefficients in the expansion of $(1 + x)^{n}$ are in the ratio $1 : 7 : 42$ . Find $n$ .

Call them $(r - 1 n)$ , $(r n)$ and $(r + 1 n)$ . The first ratio gives $\frac{n - r + 1}{r} = 7$ , so $n + 1 = 8 r$ . The second gives $\frac{n - r}{r + 1} = 6$ , so $n = 7 r + 6$ .

Substituting $n = 8 r - 1$ into the second equation gives $8 r - 1 = 7 r + 6$ , hence $r = 7$ and $n = 55$ .

Checking: $\frac{( 7 55 )}{( 6 55 )} = \frac{49}{7} = 7$ and $\frac{( 8 55 )}{( 7 55 )} = \frac{48}{8} = 6$ , and $7 \times 6 = 42$ as required.

A related question asks which terms are rational. In $(2 + 33)^{12}$ the general term carries $2^{(12 - r) /2} 3^{r /3}$ , so rationality needs $r$ divisible by $3$ and $12 - r$ even, that is $r$ divisible by $6$ . Only $r = 0, 6, 12$ qualify, so exactly three of the thirteen terms are rational.

Summary

For a positive integer index the binomial theorem is a finite identity; for any other index it is an infinite series valid only for $∣ x ∣ < 1$ , and the expression must be arranged so that its leading term is $1$ before that condition can be read. Substituting a value outside the range produces a confident and meaningless answer.

Almost every question is answered by writing $T_{r + 1} = (r n) a^{n - r} b^{r}$ , reducing the power of $x$ to a single exponent, and solving for $r$ . The middle term is one term for even $n$ and two for odd $n$ . The greatest coefficient depends only on $n$ ; the greatest term depends on $x$ as well, and is found from the ratio of consecutive terms, with equal terms when the bound is an integer.

Coefficients are extracted from products by summing a geometric series of expansions before expanding, or by absorbing negative powers into a single shifted expansion. Comparing coefficients on the two sides of $(1 + x)^{m} (1 + x)^{n} = (1 + x)^{m + n}$ proves Vandermonde's identity and, as its special case, $\sum (r n)^{2} = (n 2 n)$ .

Substituting $x = 1$ and $x = - 1$ reads off the sum of all coefficients and splits it between even and odd powers, which is why the even-indexed and odd-indexed binomial coefficients each total $2^{n - 1}$ . A given ratio of consecutive coefficients is a linear equation in $n$ and $r$ , so two ratios determine both, and a question about rational terms is a divisibility condition on $r$ read from the fractional exponents.

Sums with a plain coefficient are handled by substitution, those with a factor $r$ by differentiation or the identity $r (r n) = n (r - 1 n - 1)$ , and those with a divisor $r + 1$ by integration. For remainders, write the base as a multiple of the modulus plus or minus one, and read off the only surviving term.

Key formulas & results

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

The expansion and its general term

(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}, T_{r + 1} = (r n) a^{n - r} b^{r}

The subscript is $r+1$, so the sixth term has $r=5$. Reduce the power of $x$ to one exponent, then set it to whatever the question asks for.

Pascal's rule

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

Choosing $r$ objects either uses a fixed new object or does not. The theorem itself follows from this by induction, since multiplying by one more $(1+x)$ adds each coefficient to its neighbour.

Middle terms

n even: one, at r = \frac{n}{2}

For odd $n$ there are two middle terms, at $r=\tfrac{n-1}{2}$ and $r=\tfrac{n+1}{2}$, and they carry equal binomial coefficients.

Greatest term

\frac{T _{r + 1}}{T _{r}} = \frac{n - r + 1}{r} x > 1 ⟺ r < \frac{( n + 1 ) x}{1 + x}

Depends on $x$, unlike the greatest coefficient. If the bound is an integer, two consecutive terms are equal and both are greatest.

Greatest coefficient

(n /2 n) or (( n \pm 1 ) /2 n)

A property of $n$ alone. Consecutive coefficients rise while $r<\tfrac{n+1}{2}$, so there is exactly one peak and it sits at the middle.

Sum of all coefficients

f (1); even part = \frac{f ( 1 ) + f ( - 1 )}{2}

For $(1+x)^n$ this gives $2^{n}$ in total and $2^{n-1}$ for each parity. It also explains why $(1+x)^n+(1-x)^n$ keeps only the even powers.

Ratio of adjacent coefficients

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

A supplied ratio is a linear equation in $n$ and $r$, so two ratios determine both. Coefficients in the ratio $1:7:42$ give $n=55$ at $r=7$.

Vandermonde's identity

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

Compare coefficients of $x^{k}$ in $(1+x)^m(1+x)^n$. With $m=n=k$ it gives $\sum\binom{n}{r}^{2}=\binom{2n}{n}$.

Absorption and reversal

r (r n) = n (r - 1 n - 1); \sum f (r) = \sum f (n - r)

Choosing a committee of $r$ with a chair equals choosing the chair first. Reversing a sum and adding the two forms gives $\sum r\binom{n}{r}=n2^{n-1}$ in one line, without calculus.

Multinomial general term

\frac{n !}{n _{1} ! \dots n _{k} !} x_{1}^{n_{1}} \dots x_{k}^{n_{k}}, \sum n_{i} = n

The number of distinct terms is $\binom{n+k-1}{k-1}$ by stars and bars. For a small power it can be quicker to group two terms and use the ordinary theorem twice.

Integrated sum

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

Integrate $(1+x)^n$ from $0$ to $1$. A divisor of $r+1$ in the summand is the signal to integrate rather than to substitute.

Series for any index

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

Valid only inside that range, and only after the expression is arranged so the leading term is $1$. It terminates only when $n$ is a non-negative integer.

Negative integer index

(1 - x)^{- n} = \sum_{r \geq 0} (r n + r - 1) x^{r}

The coefficients are the stars-and-bars counts, because choosing one power from each of $n$ geometric factors is the same decision as filling $n$ boxes.

JEE Advanced test series

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

~1Est. Qs

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⚠️

Traps JEE Advanced sets — and how to dodge them

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

WATCH OUT

✗ Using a fractional or negative-index expansion outside

∣ x ∣ < 1

✓ Factor so the leading term is

1

, read off the condition, and check the given value against it.

(2 + 3 x)^{- 2}

needs

∣ x ∣ < \frac{2}{3}

Why it happens: For a positive integer index the expansion is a finite identity valid everywhere, so the habit of substituting freely is formed long before the infinite case appears.

WATCH OUT

✗ Confusing the term number with the value of

r

✓ Use

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

: the eighth term has

r = 7

Why it happens: The index starts at zero but the terms are counted from one, and the two conventions sit next to each other in the same formula.

WATCH OUT

✗ Reporting the middle term as the greatest term

✓ Compute

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

. The middle term is greatest only when

x = 1

; in

(1 + x)^{15}

with

x = \frac{1}{3}

the greatest terms are the fourth and fifth.

Why it happens: The greatest coefficient really does sit in the middle, and the two questions are phrased so similarly that the distinction is not noticed.

WATCH OUT

✗ Expanding a sum of ten binomials term by term

✓ Sum the geometric series of expansions first, then extract one coefficient from the result.

Why it happens: Each expansion looks manageable on its own, so the geometric structure of the sum is never looked for.

WATCH OUT

✗ Forgetting the sign when the general term contains a negative part

✓ Track

(- 1)^{r}

explicitly and check its parity at the value of

r

you find.

Why it happens: The magnitude calculation absorbs all the attention, and the sign is a single factor that can be dropped without the arithmetic looking wrong.

WATCH OUT

✗ Ignoring the parity of the exponent in a remainder problem

✓ When expanding

(m - 1)^{k}

, note that the surviving term is

(- 1)^{k}

, so an odd exponent leaves

- 1

rather than

1

Why it happens: The usual worked examples use

(m + 1)^{k}

, where every surviving term is

+ 1

, so the parity never has to be considered.

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 worth ~4 marks in JEE Advanced exams

5-minute revision

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

•The expansion is a finite identity only for a non-negative integer index; otherwise it is a series valid for $∣ x ∣ < 1$ .
•Arrange an expression so its leading term is $1$ before reading off the range of validity.
• $T_{r + 1}$ carries $b^{r}$ : the eighth term has $r = 7$ .
•Reduce the power of the variable to one exponent and set it to whatever the question asks for.
•The greatest coefficient depends only on $n$ ; the greatest term also depends on $x$ , through $r < \frac{( n + 1 ) x}{1 + x}$ .
•An integer value of that bound means two consecutive terms are equal and both are greatest.
•Sum a geometric series of expansions before extracting a coefficient from it.
•Comparing coefficients in $(1 + x)^{m} (1 + x)^{n}$ proves Vandermonde and, as a special case, $\sum (r n)^{2} = (n 2 n)$ .
•Multiplier $r$ means differentiate or use $r (r n) = n (r - 1 n - 1)$ ; divisor $r + 1$ means integrate.
•Reversing a sum and adding it to itself settles most linear-multiplier sums in one line.
• $f (1)$ gives the sum of all coefficients, and $\frac{f ( 1 ) \pm f ( - 1 )}{2}$ splits it by parity.
•For remainders, write the base as a multiple of the modulus plus or minus one, and watch the parity of the exponent.

JEE Advanced question blueprint

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

Typical weightage: ~1 question (roughly 3-4 marks) across the two papers combined, out of the ~120 marks of Mathematics

Question style	Marks each	Typical count	What it tests
General term, middle term and greatest term	3	1	The general term, terms independent of the variable, middle terms, and the greatest term at a specified value of the variable
Coefficients, binomial sums and identities	4	1	Coefficient extraction from products and sums of expansions, Vandermonde and related identities, and binomial sums by substitution, calculus or reversal
Fractional indices, multinomials and remainders	3	1	Expansions for negative and fractional indices with their range of validity, multinomial coefficients, rational terms, and remainders by expansion

Exam-hall strategy

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

Write the general term before reading the rest of the question. Nearly every task in this chapter is a condition on the exponent it contains.
Check the parity of $r$ whenever the expansion has a negative term. A correct magnitude with the wrong sign scores nothing.
If several expansions are being added, look for a geometric structure before expanding anything.
For any expansion with a non-integer index, write the validity condition on the same line as the expansion, not afterwards.
When a binomial sum resists, try reversing it. It costs one line and settles a large share of the standard sums.

Beyond the exam

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

Engineering approximations replace an awkward expression …

Engineering approximations replace an awkward expression by its first two binomial terms, which is why relativistic kinetic energy reduces to the familiar half-m-v-squared when the speed is far below that of light.

Compound interest and option pricing both model repeated …

Compound interest and option pricing both model repeated multiplicative steps, and the binomial distribution that results is the same array of coefficients read as probabilities.

Error-correcting codes count the number of strings within…

Error-correcting codes count the number of strings within a given distance of a codeword, which is a sum of binomial coefficients, and that sum decides how many errors the code can repair.

Where else this topic is tested

Prepare once, score in every exam that asks it.

JEE Advanced

JEE Main

BITSAT

Mathematics Olympiad (regional level)

CUET (Mathematics)

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Because the integer case terminates. Once $r$ exceeds $n$ , the factor $(n - r + 1)$ in the numerator is zero, so the series is a polynomial with $n + 1$ terms and is an identity for every $x$ . For any other index no factor ever vanishes, the series is genuinely infinite, and it equals the function only where it converges. Putting $x = 2$ into the expansion of $(1 + x)^{- 1}$ produces $1 - 2 + 4 - 8 + \dots$ , which has no sum at all.

Do not remember it; compute $\frac{( n + 1 ) x}{1 + x}$ . The middle term is greatest only when $x = 1$ , which makes that expression equal $\frac{n + 1}{2}$ . For $x$ below one the peak moves left and for $x$ above one it moves right, sometimes to the last term. The greatest coefficient is a different question and always sits at the middle.

Look at what multiplies the coefficient. Nothing means substitute a value of $x$ ; a factor of $r$ means differentiate, or use the absorption identity, or reverse the sum; a divisor of $r + 1$ means integrate. If the multiplier is quadratic in $r$ , split it as $r (r - 1) + r$ so that each piece can be absorbed separately, which is how $\sum r^{2} (r n)$ is done.

Its general term is, in the sense that questions ask for the coefficient of a specified monomial in a power of a trinomial, and the multinomial coefficient answers those directly. For small powers it is often quicker to group two of the terms and use the ordinary binomial theorem twice. The count of distinct terms, $(k - 1 n + k - 1)$ , is a stars-and-bars result rather than a new formula.

Because that expansion is a product of $n$ geometric series, and choosing one power of $x$ from each factor is exactly the decision of how many identical objects go into each of $n$ boxes. The coefficient of $x^{r}$ is therefore the stars-and-bars count $(r n + r - 1)$ . The connection is worth noticing, because it lets a counting problem be solved by extracting a coefficient and the reverse.

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Sources and How This Chapter Was CheckedSyllabus scope, what was derived rather than quoted, and how every answer here was checked.

Scope follows the JEE Advanced syllabus for 2026 (Mathematics, Binomial Theorem): the binomial theorem for a positive integral index, the general and middle terms, properties of the binomial coefficients, and simple applications.

The treatment concentrates on what Advanced adds to Main. Main asks for a named term or a coefficient in a single expansion; Advanced asks for a coefficient in a product or a sum of expansions, for the greatest term at a given value of $x$ , for a binomial sum proved by comparing coefficients, and for remainders obtained by expanding about a multiple of the modulus.

Results were derived rather than quoted. Pascal's rule came from asking whether a fixed object is chosen, the ratio test for the greatest term from dividing consecutive terms, the identity $\sum (r n)^{2} = (n 2 n)$ from comparing coefficients in a product, and the sum $\sum r^{2} (r n)$ from splitting $r^{2}$ as $r (r - 1) + r$ .

Every illustration was checked a second way. The formula for $\sum r^{2} (r n)$ was tested at $n = 2$ against a direct sum of three terms; the estimate of $1.02$ was compared with the true value to seven places; the remainder of $2^{50}$ was confirmed against the cycle of powers of $2$ modulo $7$ ; and the last two digits of $3^{400}$ were checked against the fact that $3^{4} = 81$ has order dividing $20$ modulo $100$ .

The illustrations are teaching problems written for this chapter, not previous-year questions, and are not labelled as such.

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Binomial Theorem

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

Before you start — revise these

Binomial Theorem

1. Everything from the general term

Illustration 1

Illustration 2

2. Greatest coefficient and greatest term are different questions

Illustration 3

3. Extracting a coefficient from a product

Illustration 4

Illustration 5

Illustration 6

4. Summing a binomial series

Illustration 7

5. More than two terms

Illustration 8

6. Negative and fractional indices

Illustration 9

Illustration 10

7. Divisibility and remainders

Illustration 11

Illustration 12

Illustration 13

Illustration 14

8. Reading a polynomial at x=1 and x=−1

Illustration 15

Summary

Key formulas & results

Traps JEE Advanced sets — and how to dodge them

Exam-pattern practice

Are you exam-ready for Binomial Theorem?

5-minute revision

JEE Advanced question blueprint

Exam-hall strategy

Beyond the exam

Where else this topic is tested

📝 My notes

Questions aspirants ask

Why does the expansion for a fractional index need ∣x∣<1 when the one for a positive integer index does not?

How do I remember whether the greatest term is at the middle?

When should I differentiate a binomial identity rather than substitute?

Is the multinomial theorem examinable?

Why do binomial coefficients appear in the expansion of (1−x)−n?

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8. Reading a polynomial at $x = 1$ and $x = - 1$