Binomial

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Binomial Theorem: $$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + b^n$$

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ (binomial coefficient).

Special Cases:

• $(1+x)^n = 1 + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + x^n$
• $(x-1)^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^{n-k}$

Properties of Binomial Coefficients:

• $\binom{n}{k} = \binom{n}{n-k}$ (symmetry)
• $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$ (Pascal’s identity)
• $\sum_{k=0}^{n} \binom{n}{k} = 2^n$
• $\sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0$ for $n \geq 1$

General Term: $T_{k+1} = \binom{n}{k} a^{n-k} b^k$ for $k = 0, 1, 2, \ldots, n$.

⚡ JEE Tip: For middle terms in $(a+b)^n$, when $n$ is even, the middle term is $T_{\frac{n}{2}+1}$. When $n$ is odd, two middle terms are $T_{\frac{n+1}{2}}$ and $T_{\frac{n+1}{2}+1}$.

⚡ Common Mistake: For $(1-x)^n$, the signs alternate: $T_{k+1} = \binom{n}{k}(-x)^k = (-1)^k \binom{n}{k} x^k$. Don’t forget the $(-1)^k$.

---

🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Multinomial Theorem: $(x_1 + x_2 + \cdots + x_m)^n = \sum \frac{n!}{a_1! a_2! \cdots a_m!} x_1^{a_1} x_2^{a_2} \cdots x_m^{a_m}$ where sum is over all non-negative integers $a_1, a_2, \ldots, a_m$ with $a_1 + a_2 + \cdots + a_m = n$.

Binomial Coefficient Identities:

1. $\sum_{k=0}^{n} \binom{r+k}{k} = \binom{r+n+1}{n}$ ( Hockey-stick)
2. $\sum_{k=0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}$
3. $\sum_{k=0}^{n} k^2 \binom{n}{k} = (n + n^2) 2^{n-2}$
4. $\sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}$ (special identity)

Binomial for Any Index: For any real $n$: $$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$

When $n$ is not a non-negative integer, the series is infinite and converges for $|x| < 1$.

Worked Examples:

Example 1 (JEE 2022): Find the coefficient of $x^{15}$ in $(1-x^2)^{-3}$.

Using general binomial: $(1+x)^n = \sum \binom{n}{k} x^k$ with $n = -3$. $(1-x^2)^{-3} = \sum_{k=0}^{\infty} \binom{-3}{k}(-x^2)^k = \sum_{k=0}^{\infty} \binom{-3}{k}(-1)^k x^{2k}$.

General binomial coefficient: $\binom{-3}{k} = \frac{(-3)(-4)(-5)\cdots(-3-k+1)}{k!} = \frac{(-1)^k \cdot 3 \cdot 4 \cdot 5 \cdots (k+2)}{k!} = \frac{(-1)^k (k+2)!}{2! \cdot k!} = \frac{(-1)^k (k+2)(k+1)}{2}$.

So $\binom{-3}{k}(-1)^k = \frac{(k+2)(k+1)}{2}$. Thus coefficient of $x^{2k}$ is $\frac{(k+2)(k+1)}{2}$.

We want $x^{15}$, but $x$ appears only with even powers $2k$. So coefficient of $x^{15}$ is 0.

Example 2: Find $n$ if the 3rd term in $(1+x)^n$ is 220.

The 3rd term corresponds to $k=2$: $T_3 = \binom{n}{2} x^2 \cdot 1^{n-2} = \frac{n(n-1)}{2} x^2$. Given coefficient is 220: $\frac{n(n-1)}{2} = 220$ → $n(n-1) = 440$. $n^2 - n - 440 = 0$. $(n-21)(n+20) = 0$. So $n = 21$ (positive integer).

Example 3: Find $\sum_{k=0}^{n} \binom{n}{k} 2^k$.

$(1+2)^n = 3^n = \sum_{k=0}^{n} \binom{n}{k} 1^{n-k} 2^k = \sum_{k=0}^{n} \binom{n}{k} 2^k$. So sum $= 3^n$.

---

🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious JEE Advanced preparation.

Generalised Binomial Coefficient: $$\binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$ works for any real $n$ and non-negative integer $k$.

Exponential Generating Function: The EGF for binomial expansions is related to $(1+x)^n$.

Applications in Probability: For binomial distribution: $(p+q)^n = \sum \binom{n}{k} p^k q^{n-k}$ where $p+q=1$.

Advanced Problems:

Problem (JEE Advanced 2023): Find the sum $\sum_{k=0}^{n} (-1)^k \binom{n}{k}^2$.

We need $\sum_{k=0}^{n} (-1)^k \binom{n}{k}^2$.

Using identity: $\sum_{k=0}^{n} (-1)^k \binom{n}{k}^2 = (-1)^{n/2} \binom{n}{n/2}$ if $n$ is even, and 0 if $n$ is odd.

Actually the exact result is: $\sum_{k=0}^{n} (-1)^k \binom{n}{k}^2 = \begin{cases} (-1)^{n/2} \binom{n}{n/2} & n \text{ even} \ 0 & n \text{ odd} \end{cases}$.

Proof: Using $\binom{n}{k} = \binom{n}{n-k}$ and the generating function approach. Consider $(1+x)^n (1-x)^n = (1-x^2)^n$. Coefficient of $x^n$ on left: $\sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \binom{n}{n-k} = \sum_{k=0}^n (-1)^{n-k} \binom{n}{k}^2$. Coefficient on right: 0 if $n$ is odd, $(-1)^{n/2} \binom{n}{n/2}$ if $n$ is even.

Wait, coefficient of $x^n$ in $(1-x^2)^n = \sum_{j=0}^n \binom{n}{j} (-1)^j x^{2j}$. For odd $n$, no $x^n$ term. For even $n = 2m$, coefficient of $x^{2m}$ is $(-1)^m \binom{n}{m}$. So sum $= (-1)^m \binom{2m}{m}$ when $n=2m$, and 0 when $n$ is odd.

Problem 2: Find the coefficient of $x^{10}$ in $(1-x)^{10} (1+x)^{10}$.

$((1-x)(1+x))^{10} = (1-x^2)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^{2k}$. Coefficient of $x^{10}$: since 10 is even, $2k = 10$ → $k=5$. Coefficient: $\binom{10}{5} (-1)^5 = -\binom{10}{5} = -252$.

Actually $(1-x)^{10}(1+x)^{10} = (1-x^2)^{10}$. Coefficient of $x^{10}$: when $10 = 2k$, $k=5$, coefficient $= \binom{10}{5}(-1)^5 = 252 \cdot (-1) = -252$.

JEE Advanced Patterns (2018–2024):

• General term and coefficient problems are very common
• Binomial for non-integer exponents (infinite series) appeared in 2020, 2022
• Summation of binomial coefficients with specific patterns is frequent
• Identities involving $\sum \binom{n}{k}^2$ and similar are advanced but appear
• Multinomial expansion for more than two terms is less common but tested

---

Content adapted based on your selected roadmap duration. Switch tiers using the pill selector above.