Binomial Theorem
Complete mastery of the binomial expansion, general and middle terms, greatest coefficient, multinomial theorem, and JEE-style applications including approximations and divisibility.
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01
Binomial Expansion
The binomial theorem for positive integer exponents and its fundamental properties.
The Binomial Theorem
Binomial Theorem
For any positive integer $n$:
$$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \cdots + \binom{n}{n}b^n$$
The expansion has $n+1$ terms.
Key Substitutions
$a = b = 1$: $\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n$
$a = 1, b = -1$: $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots = 0$
Adding/subtracting these:
$\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots = 2^{n-1}$ (sum of even-indexed terms)
$\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots = 2^{n-1}$ (sum of odd-indexed terms)
$a = 1, b = -1$: $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots = 0$
Adding/subtracting these:
$\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots = 2^{n-1}$ (sum of even-indexed terms)
$\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots = 2^{n-1}$ (sum of odd-indexed terms)
JEE Trick — Difference of Expansions
$(1+x)^n - (1-x)^n = 2\left[\binom{n}{1}x + \binom{n}{3}x^3 + \binom{n}{5}x^5 + \cdots\right]$
$(1+x)^n + (1-x)^n = 2\left[\binom{n}{0} + \binom{n}{2}x^2 + \binom{n}{4}x^4 + \cdots\right]$
Use these to isolate even or odd powers.
$(1+x)^n + (1-x)^n = 2\left[\binom{n}{0} + \binom{n}{2}x^2 + \binom{n}{4}x^4 + \cdots\right]$
Use these to isolate even or odd powers.
📝 Example
Find the sum $\binom{20}{0} + \binom{20}{2} + \binom{20}{4} + \cdots + \binom{20}{20}$.
This is the sum of even-indexed binomial coefficients. Answer $= 2^{20-1} = 2^{19} = \boxed{524288}$.
02
General Term & Middle Term
Finding specific terms, the general term formula, and middle term(s) — the bread and butter of JEE binomial problems.
General Term $T_{r+1}$
General Term
In the expansion of $(a+b)^n$, the $(r+1)$-th term is:
$T_{r+1} = \binom{n}{r} a^{n-r} b^r$, $\quad r = 0, 1, 2, \ldots, n$.
$T_{r+1} = \binom{n}{r} a^{n-r} b^r$, $\quad r = 0, 1, 2, \ldots, n$.
Middle Term
If $n$ is even: Middle term $= T_{n/2 + 1}$ (one middle term).
If $n$ is odd: Two middle terms $= T_{(n+1)/2}$ and $T_{(n+3)/2}$.
If $n$ is odd: Two middle terms $= T_{(n+1)/2}$ and $T_{(n+3)/2}$.
JEE Trick — Finding Independent Term
In $(x^a + x^{-b})^n$, the general term is $\binom{n}{r} x^{a(n-r) - br}$. Set the exponent $= 0$: $a(n-r) - br = 0 \Rightarrow r = \frac{an}{a+b}$. If $r$ is a non-negative integer $\leq n$, that gives the term independent of $x$.
📝 Example
Find the term independent of $x$ in $\left(x^2 + \frac{1}{x}\right)^9$.
$T_{r+1} = \binom{9}{r}(x^2)^{9-r}\left(\frac{1}{x}\right)^r = \binom{9}{r}x^{18-2r-r} = \binom{9}{r}x^{18-3r}$.
For independent term: $18 - 3r = 0 \Rightarrow r = 6$.
$T_7 = \binom{9}{6} = \boxed{84}$.
For independent term: $18 - 3r = 0 \Rightarrow r = 6$.
$T_7 = \binom{9}{6} = \boxed{84}$.
Coefficient of $x^k$ Problems
JEE Method — Finding Coefficient of $x^k$
Step 1: Write the general term $T_{r+1}$.
Step 2: Express the power of $x$ as a function of $r$.
Step 3: Set power $= k$ and solve for $r$.
Step 4: Substitute back to get the coefficient.
Step 2: Express the power of $x$ as a function of $r$.
Step 3: Set power $= k$ and solve for $r$.
Step 4: Substitute back to get the coefficient.
📝 Example
Find the coefficient of $x^{10}$ in $(1 + x)^3(1 + x^2)^7$.
$(1+x)^3 = \sum_{i=0}^{3} \binom{3}{i}x^i$. $(1+x^2)^7 = \sum_{j=0}^{7} \binom{7}{j}x^{2j}$.
Need $i + 2j = 10$ where $0 \leq i \leq 3$ and $0 \leq j \leq 7$.
$(i,j)$: $(0,5)$, $(2,4)$ are valid. ($i = 1$ gives $j = 4.5$ — not integer. $i = 3$ gives $j = 3.5$ — not integer.)
Coefficient $= \binom{3}{0}\binom{7}{5} + \binom{3}{2}\binom{7}{4} = 1 \times 21 + 3 \times 35 = 21 + 105 = \boxed{126}$.
Need $i + 2j = 10$ where $0 \leq i \leq 3$ and $0 \leq j \leq 7$.
$(i,j)$: $(0,5)$, $(2,4)$ are valid. ($i = 1$ gives $j = 4.5$ — not integer. $i = 3$ gives $j = 3.5$ — not integer.)
Coefficient $= \binom{3}{0}\binom{7}{5} + \binom{3}{2}\binom{7}{4} = 1 \times 21 + 3 \times 35 = 21 + 105 = \boxed{126}$.
03
Greatest Coefficient & Greatest Term
Identifying the numerically largest coefficient and the numerically largest term in a binomial expansion.
Greatest Binomial Coefficient
Greatest Coefficient in $(1+x)^n$
The greatest binomial coefficient $\binom{n}{r}$ is:
$n$ even: $\binom{n}{n/2}$ (the middle coefficient).
$n$ odd: $\binom{n}{(n-1)/2} = \binom{n}{(n+1)/2}$ (two equal middle coefficients).
$n$ even: $\binom{n}{n/2}$ (the middle coefficient).
$n$ odd: $\binom{n}{(n-1)/2} = \binom{n}{(n+1)/2}$ (two equal middle coefficients).
Greatest Term in $(a+b)^n$
To find the numerically greatest term, consider the ratio:
$\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \cdot \frac{b}{a}$
$T_{r+1} \geq T_r$ when $\frac{n-r+1}{r} \cdot \left|\frac{b}{a}\right| \geq 1$, i.e., $r \leq \frac{(n+1)|b|}{|a|+|b|}$.
JEE Trick: Let $m = \frac{(n+1)|b|}{|a|+|b|}$. If $m$ is an integer, $T_m$ and $T_{m+1}$ are both greatest. Otherwise, $T_{\lfloor m \rfloor + 1}$ is the greatest term.
$\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \cdot \frac{b}{a}$
$T_{r+1} \geq T_r$ when $\frac{n-r+1}{r} \cdot \left|\frac{b}{a}\right| \geq 1$, i.e., $r \leq \frac{(n+1)|b|}{|a|+|b|}$.
JEE Trick: Let $m = \frac{(n+1)|b|}{|a|+|b|}$. If $m$ is an integer, $T_m$ and $T_{m+1}$ are both greatest. Otherwise, $T_{\lfloor m \rfloor + 1}$ is the greatest term.
📝 Example
Find the greatest term in $(2 + 3x)^9$ when $x = 1$.
Here $a = 2$, $b = 3$, $n = 9$. $m = \frac{(9+1) \times 3}{2+3} = \frac{30}{5} = 6$.
Since $m = 6$ is an integer, both $T_6$ and $T_7$ are greatest terms.
$T_7 = \binom{9}{6}(2)^3(3)^6 = 84 \times 8 \times 729 = \boxed{489888}$.
Since $m = 6$ is an integer, both $T_6$ and $T_7$ are greatest terms.
$T_7 = \binom{9}{6}(2)^3(3)^6 = 84 \times 8 \times 729 = \boxed{489888}$.
04
Multinomial Theorem
Extension to more than two terms — multinomial expansion and its JEE applications.
Multinomial Expansion
Multinomial Theorem
$(x_1 + x_2 + \cdots + x_k)^n = \sum \frac{n!}{n_1! n_2! \cdots n_k!} x_1^{n_1} x_2^{n_2} \cdots x_k^{n_k}$
where the sum is over all non-negative integers $n_1, n_2, \ldots, n_k$ with $n_1 + n_2 + \cdots + n_k = n$.
where the sum is over all non-negative integers $n_1, n_2, \ldots, n_k$ with $n_1 + n_2 + \cdots + n_k = n$.
Number of Terms
$(x_1 + x_2 + \cdots + x_k)^n$ has $\binom{n+k-1}{k-1}$ terms.
JEE Shortcut: Number of terms in $(1 + x + x^2)^n$ considering distinct powers of $x$: the maximum power is $2n$, so there are $2n + 1$ distinct terms.
JEE Shortcut: Number of terms in $(1 + x + x^2)^n$ considering distinct powers of $x$: the maximum power is $2n$, so there are $2n + 1$ distinct terms.
📝 Example
Find the coefficient of $x^3y^2z$ in $(x + y + z)^6$.
By multinomial theorem: $\frac{6!}{3! \cdot 2! \cdot 1!} = \frac{720}{6 \times 2 \times 1} = \boxed{60}$.
05
Applications — Approximations, Divisibility & Summation
Using binomial theorem for approximations, divisibility proofs, sum of coefficients, and summation of series — JEE Advanced staples.
Approximations & Divisibility
Binomial Approximation
For $|x| \ll 1$: $(1+x)^n \approx 1 + nx$ (first-order approximation).
More precisely: $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots$
JEE Use: To approximate expressions like $(1.01)^{10}$, $(0.99)^5$, etc.
More precisely: $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots$
JEE Use: To approximate expressions like $(1.01)^{10}$, $(0.99)^5$, etc.
Divisibility Problems
To show $a^n - b^n$ is divisible by $(a-b)$: write $a = b + (a-b)$ and expand using binomial theorem.
Common JEE pattern: To find the remainder when $a^n$ is divided by $m$, write $a = m \cdot q + r$ and expand $(mq + r)^n$.
Common JEE pattern: To find the remainder when $a^n$ is divided by $m$, write $a = m \cdot q + r$ and expand $(mq + r)^n$.
JEE Trick — Sum of Coefficients
Sum of coefficients of $f(x) = f(1)$.
Sum of coefficients of $(1 + x + x^2)^n = 3^n$ (put $x = 1$).
Sum of coefficients of $(1 - 3x + 5x^2)^{10}$: put $x = 1$ to get $(1-3+5)^{10} = 3^{10}$.
Sum of coefficients of $(1 + x + x^2)^n = 3^n$ (put $x = 1$).
Sum of coefficients of $(1 - 3x + 5x^2)^{10}$: put $x = 1$ to get $(1-3+5)^{10} = 3^{10}$.
📝 Example
Find the remainder when $7^{82}$ is divided by $6$.
$7 = 6 + 1$. So $7^{82} = (6+1)^{82} = \sum_{r=0}^{82} \binom{82}{r} 6^r \cdot 1^{82-r}$.
All terms with $r \geq 1$ are divisible by $6$. The term with $r = 0$ is $\binom{82}{0} \cdot 1 = 1$.
Remainder $= \boxed{1}$.
All terms with $r \geq 1$ are divisible by $6$. The term with $r = 0$ is $\binom{82}{0} \cdot 1 = 1$.
Remainder $= \boxed{1}$.
Summation of Binomial Series
Differentiation & Integration Method
Differentiation: Differentiate $(1+x)^n = \sum \binom{n}{r}x^r$ to get:
$n(1+x)^{n-1} = \sum r\binom{n}{r}x^{r-1}$. Put $x = 1$: $\sum r\binom{n}{r} = n \cdot 2^{n-1}$.
Integration: Integrate $(1+x)^n$ from $0$ to $1$:
$\frac{2^{n+1}-1}{n+1} = \sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1}$.
$n(1+x)^{n-1} = \sum r\binom{n}{r}x^{r-1}$. Put $x = 1$: $\sum r\binom{n}{r} = n \cdot 2^{n-1}$.
Integration: Integrate $(1+x)^n$ from $0$ to $1$:
$\frac{2^{n+1}-1}{n+1} = \sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1}$.
JEE Trick — Telescoping with Binomials
$\sum_{r=0}^{n} r^2 \binom{n}{r} = n(n-1)2^{n-2} + n \cdot 2^{n-1}$ (differentiate twice).
$\sum_{r=0}^{n} \frac{(-1)^r}{r+1}\binom{n}{r} = \frac{1}{n+1}$ (integrate from $0$ to $1$ with alternating signs).
$\sum_{r=0}^{n} \frac{(-1)^r}{r+1}\binom{n}{r} = \frac{1}{n+1}$ (integrate from $0$ to $1$ with alternating signs).
📝 Example
Evaluate: $\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \cdots + \binom{n}{n}^2$.
Consider $(1+x)^n(1+x)^n = (1+x)^{2n}$. Comparing the coefficient of $x^n$ on both sides:
LHS: $\sum_{r=0}^{n} \binom{n}{r}\binom{n}{n-r} = \sum_{r=0}^{n} \binom{n}{r}^2$.
RHS: $\binom{2n}{n}$.
So $\sum_{r=0}^{n}\binom{n}{r}^2 = \boxed{\binom{2n}{n}}$. (Vandermonde's identity.)
LHS: $\sum_{r=0}^{n} \binom{n}{r}\binom{n}{n-r} = \sum_{r=0}^{n} \binom{n}{r}^2$.
RHS: $\binom{2n}{n}$.
So $\sum_{r=0}^{n}\binom{n}{r}^2 = \boxed{\binom{2n}{n}}$. (Vandermonde's identity.)
★ Key Takeaways
- The general term is $T_{r+1} = \binom{n}{r}a^{n-r}b^r$ — memorise this form and practise finding specific terms quickly.
- For the term independent of $x$, set the power of $x$ to zero in the general term.
- Sum of coefficients $=$ value of the expression at $x = 1$. Sum of coefficients with alternating signs $=$ value at $x = -1$.
- Divisibility and remainder problems: write the base as $(m + r)$ and expand — all terms except the last are divisible by $m$.
- Differentiation and integration of $(1+x)^n$ generate powerful sum identities — master these for JEE Advanced.
✎ Practice Problems
Problem 1
Find the coefficient of $x^5$ in $(1 + x)^{11}(1 - x + x^2)^{11}$.
Show Solution ▼
$(1+x)(1 - x + x^2) = 1 + x^3$. So $(1+x)^{11}(1-x+x^2)^{11} = (1+x^3)^{11}$.
General term: $\binom{11}{r}x^{3r}$. For $x^5$: $3r = 5 \Rightarrow r = 5/3$ — not integer. Coefficient of $x^5 = \boxed{0}$.
General term: $\binom{11}{r}x^{3r}$. For $x^5$: $3r = 5 \Rightarrow r = 5/3$ — not integer. Coefficient of $x^5 = \boxed{0}$.
Problem 2
Find the middle term of $\left(\frac{x}{y} + \frac{y}{x}\right)^{10}$.
Show Solution ▼
$n = 10$ (even), so middle term $= T_6$.
$T_6 = \binom{10}{5}\left(\frac{x}{y}\right)^5\left(\frac{y}{x}\right)^5 = \binom{10}{5} \cdot 1 = \boxed{252}$.
$T_6 = \binom{10}{5}\left(\frac{x}{y}\right)^5\left(\frac{y}{x}\right)^5 = \binom{10}{5} \cdot 1 = \boxed{252}$.
Problem 3
If $\binom{n}{4}$, $\binom{n}{5}$, and $\binom{n}{6}$ are in AP, find $n$.
Show Solution ▼
AP condition: $2\binom{n}{5} = \binom{n}{4} + \binom{n}{6}$.
$\frac{\binom{n}{5}}{\binom{n}{4}} = \frac{n-4}{5}$ and $\frac{\binom{n}{6}}{\binom{n}{5}} = \frac{n-5}{6}$.
Dividing the AP condition by $\binom{n}{5}$: $2 = \frac{5}{n-4} + \frac{n-5}{6}$.
$2 = \frac{5}{n-4} + \frac{n-5}{6}$. Let $m = n-4$: $2 = \frac{5}{m} + \frac{m-1}{6}$.
$12m = 30 + m(m-1) = m^2 - m + 30$. $m^2 - 13m + 30 = 0$. $(m-3)(m-10) = 0$.
$m = 3 \Rightarrow n = 7$ or $m = 10 \Rightarrow n = 14$. Both valid since $n \geq 6$. Answer: $n = 7$ or $n = 14$.
$\frac{\binom{n}{5}}{\binom{n}{4}} = \frac{n-4}{5}$ and $\frac{\binom{n}{6}}{\binom{n}{5}} = \frac{n-5}{6}$.
Dividing the AP condition by $\binom{n}{5}$: $2 = \frac{5}{n-4} + \frac{n-5}{6}$.
$2 = \frac{5}{n-4} + \frac{n-5}{6}$. Let $m = n-4$: $2 = \frac{5}{m} + \frac{m-1}{6}$.
$12m = 30 + m(m-1) = m^2 - m + 30$. $m^2 - 13m + 30 = 0$. $(m-3)(m-10) = 0$.
$m = 3 \Rightarrow n = 7$ or $m = 10 \Rightarrow n = 14$. Both valid since $n \geq 6$. Answer: $n = 7$ or $n = 14$.
Problem 4
Find the last two digits of $3^{100}$ (i.e., $3^{100} \mod 100$).
Show Solution ▼
$3^{100} = (3^2)^{50} = 9^{50} = (10-1)^{50} = \sum_{r=0}^{50}\binom{50}{r}10^r(-1)^{50-r}$.
For last two digits, only $r = 0$ and $r = 1$ matter (higher terms are divisible by $100$):
$r=0$: $(-1)^{50} = 1$. $r=1$: $\binom{50}{1} \cdot 10 \cdot (-1)^{49} = -500 \equiv 0 \pmod{100}$.
Wait — need $r=0$ and $r=1$: $1 + (-500) = -499 \equiv 01 \pmod{100}$. Last two digits: $\boxed{01}$.
For last two digits, only $r = 0$ and $r = 1$ matter (higher terms are divisible by $100$):
$r=0$: $(-1)^{50} = 1$. $r=1$: $\binom{50}{1} \cdot 10 \cdot (-1)^{49} = -500 \equiv 0 \pmod{100}$.
Wait — need $r=0$ and $r=1$: $1 + (-500) = -499 \equiv 01 \pmod{100}$. Last two digits: $\boxed{01}$.
Problem 5
Find the value of $\sum_{r=0}^{10} r \cdot \binom{10}{r} \cdot 3^r$.
Show Solution ▼
We know $\sum_{r=0}^{n}r\binom{n}{r}x^r = nx(1+x)^{n-1}$ (differentiate $(1+x)^n = \sum \binom{n}{r}x^r$ and multiply by $x$).
Put $n = 10$, $x = 3$: $10 \times 3 \times 4^9 = 30 \times 4^9 = 30 \times 262144 = \boxed{7864320}$.
Put $n = 10$, $x = 3$: $10 \times 3 \times 4^9 = 30 \times 4^9 = 30 \times 262144 = \boxed{7864320}$.
🎯 Interactive MCQs
1. The coefficient of $x^4$ in $\left(1 + x - 2x^2\right)^7$ is:
A $91$
B $-91$
C $35$
D $-35$
2. If the coefficients of three consecutive terms in $(1+x)^n$ are in the ratio $1:7:42$, then $n$ equals:
A $49$
B $55$
C $42$
D $63$
3. The number of rational terms in the expansion of $\left(\sqrt{2} + \sqrt[3]{3}\right)^{12}$ is:
A $2$
B $3$
C $4$
D $5$
4. The value of $\sum_{r=0}^{n} (-1)^r \binom{n}{r}\left(\frac{1}{2^r}\right)$ is:
A $\frac{1}{2^n}$
B $\frac{1}{3^n}$
C $0$
D $2^n$
5. The integral part of $(5\sqrt{5} + 11)^{2n+1}$ is:
A An even number
B An odd number
C Divisible by $5$
D A prime number