Mathematics Binomial Theorem

Binomial Theorem Formula Sheet

All key Binomial Theorem formulas — expansion, general & middle term, coefficient identities, multinomial theorem and approximations for JEE quick revision.

6 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every must-know Binomial Theorem result in one scannable place — expansion, general and middle term, coefficient identities, the multinomial theorem, and approximation formulas. Use this for last-minute JEE Main and Advanced revision.

The Binomial Theorem

For any positive integer $n$:

$$\boxed{(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r}$$$$= \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n$$

where the binomial coefficient is

$$\binom{n}{r} = {}^nC_r = \frac{n!}{r!(n-r)!}$$
Core facts

The expansion of $(a+b)^n$ has exactly $(n+1)$ terms, and in every term the sum of the powers of $a$ and $b$ equals $n$.

Standard forms

FormExpansion
$(1 + x)^n$$1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \cdots + x^n$
$(1 - x)^n$$1 - nx + \dfrac{n(n-1)}{2!}x^2 - \cdots + (-1)^n x^n$
$(x + a)^n$$x^n + nx^{n-1}a + \dfrac{n(n-1)}{2!}x^{n-2}a^2 + \cdots + a^n$
Sign trap in $(a-b)^n$

For $(a - b)^n$, replace $b$ with $(-b)$ and include the sign before raising to the power. E.g. in $(x-2y)^5$, $T_3 = \binom{5}{2}x^3(-2y)^2 = 40x^3y^2$, not $\binom{5}{2}x^3(2y)^2$.

General Term & Term Finding

The general term — the $(r+1)^{\text{th}}$ term — of $(a+b)^n$:

$$\boxed{T_{r+1} = \binom{n}{r} a^{n-r} b^r}, \quad r = 0, 1, 2, \ldots, n$$
ExpansionGeneral term $T_{r+1}$
$(x + a)^n$$\binom{n}{r} x^{n-r} a^r$
$(1 + x)^n$$\binom{n}{r} x^r$
$(ax + b)^n$$\binom{n}{r} a^{n-r} b^r x^{n-r}$
$\left(x + \tfrac{1}{x}\right)^n$$\binom{n}{r} x^{n-2r}$
r vs term number

Term number $= r + 1$. The 5th term means $T_5$, so $r = 4$. Term independent of $x$: set the net power of $x$ in $T_{r+1}$ to $0$ and solve for $r$. If $r$ is not a non-negative integer, that term (or coefficient) is $0$.

Term from the end

$$r^{\text{th}} \text{ term from the end} = (n - r + 2)^{\text{th}} \text{ term from the start}$$

Middle Term(s)

CaseMiddle term(s)
$n$ evenone term: $T_{\frac{n}{2}+1}$
$n$ oddtwo terms: $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$

For $n$ even:

$$\boxed{T_{\frac{n}{2}+1} = \binom{n}{n/2}\, a^{n/2} b^{n/2}}$$

For $n$ odd:

$$\boxed{T_{\frac{n+1}{2}} = \binom{n}{(n-1)/2}\, a^{(n+1)/2} b^{(n-1)/2}}$$$$\boxed{T_{\frac{n+3}{2}} = \binom{n}{(n+1)/2}\, a^{(n-1)/2} b^{(n+1)/2}}$$

Greatest coefficient: $\binom{n}{n/2}$ if $n$ is even; $\binom{n}{(n-1)/2} = \binom{n}{(n+1)/2}$ if $n$ is odd.

Greatest Term (Numerically)

For $(a + b)^n$, the numerically greatest term is $T_{r+1}$ where

$$\boxed{r = \left[\frac{(n+1)\,|b|}{|a| + |b|}\right]} \quad ([\,\cdot\,] = \text{greatest integer})$$

Equivalently, $T_{r+1} > T_r$ as long as

$$\frac{(n - r + 1)\,|b|}{r\,|a|} > 1$$

For $(1+x)^n$ with $x > 0$: if $\dfrac{(n+1)x}{1+x}$ is an integer $m$, then $T_m$ and $T_{m+1}$ are both greatest; otherwise $T_{[m]+1}$ is greatest.

Binomial Coefficient Properties

$$\binom{n}{0} = \binom{n}{n} = 1, \quad \binom{n}{1} = \binom{n}{n-1} = n$$
IdentityFormula
Symmetry$\binom{n}{r} = \binom{n}{n-r}$
Pascal’s identity$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$
Sum of consecutive$\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$
Ratio of consecutive$\dfrac{\binom{n}{r}}{\binom{n}{r-1}} = \dfrac{n-r+1}{r}$
Index relation$r\binom{n}{r} = n\binom{n-1}{r-1}$
Symmetry needs equal top index

$\binom{10}{3} = \binom{10}{7}$, not $\binom{3}{10}$. And Pascal’s identity goes to the previous row: $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$ — never same-row neighbours.

Sum Identities (with $C_r = \binom{n}{r}$)

SumResult
$C_0 + C_1 + C_2 + \cdots + C_n$$2^n$
$C_0 - C_1 + C_2 - \cdots + (-1)^n C_n$$0$
$C_0 + C_2 + C_4 + \cdots$$2^{n-1}$
$C_1 + C_3 + C_5 + \cdots$$2^{n-1}$
$C_1 + 2C_2 + 3C_3 + \cdots + nC_n = \sum r\,C_r$$n\cdot 2^{n-1}$
$\sum_{r=0}^{n} r^2 \binom{n}{r}$$n(n+1)\,2^{n-2}$
$C_0^2 + C_1^2 + \cdots + C_n^2 = \sum \binom{n}{r}^2$$\binom{2n}{n}$

Headline results:

$$\boxed{\sum_{r=0}^{n}\binom{n}{r} = 2^n} \qquad \boxed{\sum_{r=0}^{n}\binom{n}{r}^2 = \binom{2n}{n}}$$$$\boxed{\sum_{r=0}^{n} r\binom{n}{r} = n\cdot 2^{n-1}}$$

Vandermonde’s identity

$$\boxed{\sum_{r=0}^{k} \binom{m}{r}\binom{n}{k-r} = \binom{m+n}{k}}$$

Integration / differentiation results

$$\sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1} = \frac{2^{n+1} - 1}{n+1}$$$$\sum_{r=0}^{n}(-1)^r\binom{n}{r}^2 = \begin{cases} 0 & n \text{ odd} \\ (-1)^{n/2}\binom{n}{n/2} & n \text{ even} \end{cases}$$
How sum identities are generated

Put $x = 1$ in $(1+x)^n$ for the total $2^n$; put $x = -1$ for the alternating sum $0$. For weighted sums ($\sum rC_r$, $\sum r^2 C_r$) differentiate $(1+x)^n$ then put $x=1$; for $\sum \frac{C_r}{r+1}$ integrate from $0$ to $1$.

Multinomial Theorem

For a positive integer $n$ and $k$ variables:

$$\boxed{(x_1 + x_2 + \cdots + x_k)^n = \sum \frac{n!}{n_1!\, n_2! \cdots n_k!}\, x_1^{n_1} x_2^{n_2} \cdots x_k^{n_k}}$$

where $n_1 + n_2 + \cdots + n_k = n$, all $n_i \ge 0$.

Multinomial coefficient:

$$\binom{n}{n_1, n_2, \ldots, n_k} = \frac{n!}{n_1!\, n_2! \cdots n_k!} = \binom{n}{n_1}\binom{n-n_1}{n_2}\binom{n-n_1-n_2}{n_3}\cdots$$

Trinomial special case:

$$(a + b + c)^n = \sum_{p+q+r=n} \frac{n!}{p!\,q!\,r!}\, a^p b^q c^r$$
QuantityFormulaNotes
Number of terms$\binom{n+k-1}{k-1} = \binom{n+k-1}{n}$stars-and-bars; $\binom{n+2}{2}$ for trinomial
Sum of all coefficients$k^n$put every variable $= 1$
Greatest coefficientpowers as equal as possibleif $n = kq + r$: $r$ terms get power $q+1$, rest get $q$
Factorials go on the bottom

Coefficient of $x^a y^b z^c$ in $(x+y+z)^n$ is $\dfrac{n!}{a!\,b!\,c!}$ — total factorial on top. It exists only if $a + b + c = n$; otherwise the coefficient is $0$.

Approximations (small $x$)

For $|x| < 1$ and any real $n$ (positive, negative, or fractional):

$$\boxed{(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots}$$
  • First order: $(1+x)^n \approx 1 + nx$
  • Second order: $(1+x)^n \approx 1 + nx + \dfrac{n(n-1)}{2}x^2$
ExpressionApproximation (small $x$)
$(1 + x)^n$$1 + nx$
$(1 - x)^n$$1 - nx$
$\sqrt{1 + x}$$1 + \dfrac{x}{2}$
$\sqrt[3]{1 + x}$$1 + \dfrac{x}{3}$
$\dfrac{1}{1 + x}$$1 - x$
$\dfrac{1}{\sqrt{1 + x}}$$1 - \dfrac{x}{2}$
$\dfrac{1}{(1 + x)^2}$$1 - 2x$

Higher-order single-variable expansions used in JEE:

$$\sqrt{1+x} \approx 1 + \frac{x}{2} - \frac{x^2}{8}, \qquad \sqrt[3]{1+x} \approx 1 + \frac{x}{3} - \frac{x^2}{9}$$$$\frac{1}{1+x} = (1+x)^{-1} \approx 1 - x + x^2 - x^3 + \cdots$$

Relative error when using $(1+x)^n \approx 1 + nx$:

$$\text{Relative error} \approx \frac{n(n-1)x^2}{2}$$
When approximation is valid

Use $(1+x)^n \approx 1+nx$ only when $|x| \ll 1$ (typically $|x| < 0.1$). To apply it, first factor the base close to $1$: e.g. $(2.03)^5 = 2^5\left(1 + \tfrac{0.03}{2}\right)^5$ and $\sqrt{99} = 10\left(1 - \tfrac{1}{100}\right)^{1/2}$.

Pascal’s Triangle

Each entry is the sum of the two directly above it (Pascal’s identity), giving the binomial coefficients row by row.

graph TD
    A["n=0:  1"] --> B["n=1:  1  1"]
    B --> C["n=2:  1  2  1"]
    C --> D["n=3:  1  3  3  1"]
    D --> E["n=4:  1  4  6  4  1"]
    E --> F["n=5:  1  5  10  10  5  1"]

One-Glance Summary

QuantityFormula
Expansion$(a+b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
Number of terms$n + 1$
General term$T_{r+1} = \binom{n}{r}a^{n-r}b^r$
Middle term ($n$ even)$T_{\frac{n}{2}+1}$
Middle terms ($n$ odd)$T_{\frac{n+1}{2}},\ T_{\frac{n+3}{2}}$
Greatest term index$r = \left[\dfrac{(n+1)
Symmetry$\binom{n}{r} = \binom{n}{n-r}$
Pascal’s identity$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$
Ratio$\dfrac{\binom{n}{r}}{\binom{n}{r-1}} = \dfrac{n-r+1}{r}$
$\sum C_r$$2^n$
$\sum (-1)^r C_r$$0$
Even = odd place sum$2^{n-1}$
$\sum r\,C_r$$n\,2^{n-1}$
$\sum r^2 C_r$$n(n+1)2^{n-2}$
$\sum C_r^2$$\binom{2n}{n}$
Vandermonde$\sum_r \binom{m}{r}\binom{n}{k-r} = \binom{m+n}{k}$
Multinomial terms$\binom{n+k-1}{k-1}$
Multinomial coeff. sum$k^n$
Approximation$(1+x)^n \approx 1 + nx,\