Binomial Theorem — Complete JEE Mains 2026 Revision Capsule
This capsule collects every formula and identity you need for JEE Mains: finite binomial theorem, coefficients, combinatorial identities, Newton’s generalized binomial, approximations, problem patterns, and practice questions — all equations in LaTeX.
1. Finite Binomial Theorem (Standard)
For any integer $n\\ge 0$, $$ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{\,n-k} y^{\,k} $$ where $$ \binom{n}{k} = \frac{n!}{k!(n-k)!}. $$
General term (T_{k+1}):
$$ T_{k+1} = \binom{n}{k} x^{\,n-k} y^{\,k}, \qquad k = 0,1,\dots,n. $$
Special cases:
- $(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$
- Coefficient of $x^r$ in $(1+x)^n$ is $\binom{n}{r}$.
- Sum of coefficients: $(1+1)^n = 2^n$.
- Alternating sum: $(1-1)^n = 0$ for $n\\ge 1$ (useful for cancellations).
2. Binomial Coefficient Identities & Pascal
Pascal identity:
$$ \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}. $$
Symmetry:
$$ \binom{n}{k} = \binom{n}{n-k}. $$
Summation identities:
- $$ \sum_{k=0}^n \binom{n}{k} = 2^n. $$
- $$ \sum_{k=0}^n (-1)^k \binom{n}{k} = 0 \quad (n\ge1). $$
- $$ \sum_{k=0}^n k\binom{n}{k} = n2^{n-1}. $$
- $$ \sum_{k=0}^n k^2\binom{n}{k} = n(n+1)2^{n-2}. $$
- $$ \sum_{k=0}^n \binom{r+k}{k} = \binom{r+n+1}{n} \quad (\text{hockey-stick identity}). $$
Combinatorial meaning: $\binom{n}{k}$ = number of ways to choose $k$ items from $n$ (order irrelevant).
3. Useful Properties & Relations
- $$ \binom{n}{0} = \binom{n}{n} = 1. $$
- $$ \binom{n}{1} = \binom{n}{n-1} = n. $$
- $$ \binom{n}{k} = \frac{n}{k}\binom{n-1}{k-1}. $$
- For integer $k>n$, $\binom{n}{k}=0$ (conventionally).
- $$ \binom{n}{k}\binom{k}{r} = \binom{n}{r}\binom{n-r}{k-r}. $$
4. Multinomial Theorem (Brief)
For $n\\ge0$ and variables $x_1,\\dots,x_m$, $$ (x_1 + x_2 + \dots + x_m)^n = \sum_{k_1+\\dots+k_m = n} \frac{n!}{k_1!k_2!\\cdots k_m!} \prod_{i=1}^m x_i^{k_i}. $$
Multinomial coefficients reduce to binomial when $m=2$.
5. Coefficient Extraction & Tricks
Coefficient of $x^r$ in $(ax + b)^n$:
$$ [x^r](ax + b)^n = \binom{n}{r} a^r b^{n-r}. $$
Coefficient in product expansions: Use convolution: if $(1+x)^n(1+x)^m = (1+x)^{n+m}$, coefficients add via convolution.
Finding particular coefficient: Replace $x$ by $x/y$ etc. — common trick: to get coefficient of $x^r$ in $(1+2x)^n(1+3x)^m$ expand each or use generating functions / convolution.
6. Newton’s Generalized Binomial Theorem (Non-integer exponents)
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First few terms:
$$ (1+x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3 + \cdots $$
Convergence: the series converges for $|x|<1 a="" alpha="" binomial="" finite="" for="" integer="" it="" non-negative="" p="" terminates="">
Special expansions (useful):
- $(1+x)^{-1} = 1 - x + x^2 - x^3 + \\cdots$ for $|x|<1 li="">
- $(1+x)^{1/2} = 1 + \\tfrac{1}{2}x - \\tfrac{1}{8}x^2 + \\tfrac{1}{16}x^3 + \\cdots$ for $|x|<1 li="">
7. Approximations, Inequalities & Useful Bounds
- Binomial bounds: For $0 \\le k \\le n$, $$ \binom{n}{k} \\le \frac{n^k}{k!}. $$ Useful when estimating sizes.
- Dominant term idea: For $(1+x)^n$ with $n$ large and small $x$, main contribution near $k\\approx nx/(1+x)$ (useful for combinatorial heuristics).
- Using binomial to approximate $(1+x)^n$: Keep first two/three terms when $x$ small and $n$ moderate.
8. JEE Traps & Shortcuts
- Trap: forgetting symmetry — coefficient of $x^r$ equals coefficient of $x^{n-r}$ in $(1+x)^n$ because $\binom{n}{r} = \binom{n}{n-r}$.
- Trap: sign errors in alternating expansions: $(1-1)^n = 0$ cancels everything — but if replaced by $(1-1)^n f(n)$ watch where $f$ depends on $k$.
- Shortcut: use $(1+1)^n$ to get sum of coefficients; use derivative w.r.t $x$ on $(1+x)^n$ to get sums of $k\binom{n}{k}$ etc: $$ \frac{d}{dx}(1+x)^n = n(1+x)^{n-1} = \sum_{k=0}^n k\binom{n}{k}x^{k-1}. $$
- Trap: coefficient of $x^r$ in product of binomials sometimes easier via generating functions rather than full expansion.
- Tip: For coefficient of $x^m$ in $(ax+ b)^n$, factor out $b^n$ and convert: coefficient = $b^{n-m}a^m\binom{n}{m}$ after appropriate scaling.
9. Worked Examples (High-yield)
Example 1: Find coefficient of $x^{5}$ in $(1+x)^{10}$.
Solution: Coefficient = $\binom{10}{5} = 252$.
Example 2: Coefficient of $x^3$ in $(2x-3)^6$.
Solution: Use general term: $T_{k+1} = \binom{6}{k} (2x)^{6-k}(-3)^k$. For $x^3$ we need $6-k = 3 \\Rightarrow k=3$. Coefficient = $\binom{6}{3} 2^3 (-3)^3 = 20\\cdot8\\cdot(-27) = -4320$.
Example 3 (Sum using binomial): Evaluate $\sum_{k=0}^{n} k\binom{n}{k}$.
Solution: Using derivative method at $x=1$: $\sum k\binom{n}{k} = n2^{n-1}$.
10. Practice Problems (Try first)
- Find coefficient of $x^4$ in $(1+2x)^8$.
- Find coefficient of $x^7$ in $(x+1)^{15}$. (Quick)
- Evaluate $\sum_{k=0}^{n} (-1)^k \binom{n}{k} k^2$.
- Expand $(1+x)^{1/2}$ up to $x^3$ term using Newton’s generalized theorem.
- Find constant term in $(x - \\frac{1}{x})^{10}$.
- Show that $\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$.
Show Answers & Hints
- Coefficient = $\binom{8}{4} 2^4 = 70\\cdot16 = 1120$.
- Coefficient = $\binom{15}{7} = 6435$.
- Use generating function and derivative: $\sum (-1)^k k^2\binom{n}{k}$ equals $n2^{n-2}(n+1)$ times a sign — easier: compute via $(1-1)^n$ derivatives; final result = $0$ for first derivative-based sums, but compute directly: $\sum (-1)^k k\binom{n}{k} = 0$ and $\sum (-1)^k k^2\binom{n}{k} = 0$ for $n\\ge2$ (use identities).
- $(1+x)^{1/2} = 1 + \\tfrac{1}{2}x - \\tfrac{1}{8}x^2 + \\tfrac{1}{16}x^3 + \\cdots$ so up to $x^3$ keep first four terms.
- Constant term in $(x - 1/x)^{10}$: Expand binomial: term with $x^{10-2k}$ and set exponent 0 → $10-2k=0 \\Rightarrow k=5$. Coefficient = $\binom{10}{5}(-1)^5 = -252$.
- Use coefficient of $x^n$ in $(1+x)^{2n}$ equals $\sum_{k=0}^n \binom{n}{k}\binom{n}{n-k} = \sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$.
11. One-line Takeaways
- Finite binomial theorem for integer exponents; Newton’s generalized for real exponents with $|x|<1 li="">
- Use Pascal identity & symmetry for coefficient manipulations.
- Derivative of generating function yields sums involving $k, k^2,$ etc.
- Telescoping / convolution are your friends for product expansions and sums.
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