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The binomial theorem is one of the most predictable chapters in JEE maths. Questions ask for a coefficient, the term independent of x, a middle term or a sum of coefficients, and all of them come from the same general term. The marks are lost in small places: a missing minus sign, r confused with r + 1, or a power forgotten on a coefficient. This post works through each question type with checked arithmetic.
What does the syllabus cover?
The JEE (Main) 2026 syllabus lists the "binomial theorem for a positive integral index, general term and middle term and simple applications". The JEE (Advanced) 2026 syllabus adds properties of binomial coefficients. NCERT's Class 11 chapter on the binomial theorem covers the theorem, its proof and Pascal's triangle. The CBSE 2026-27 curriculum lists "general and middle term in binomial expansion" among Class 11 topics that will be assessed only formatively, so school tests may not ask for them, but JEE does. Check the documents for your exam year.
What is the general term?
For (a + b)ⁿ, the (r + 1)th term is
T(r+1) = ⁿCᵣ · aⁿ⁻ʳ · bʳ, for r = 0, 1, …, n.
There are n + 1 terms. The power of b equals r, which is one less than the term's position. Most errors come from losing track of that.
How do you find a particular coefficient?
Example. Find the coefficient of x⁵ in (2x − 3)⁸.
General term: ⁸Cᵣ (2x)⁸⁻ʳ (−3)ʳ.
The power of x is 8 − r. Set 8 − r = 5, so r = 3.
Coefficient: ⁸C₃ × 2⁵ × (−3)³ = 56 × 32 × (−27).
56 × 32 = 1792, and 1792 × 27 = 48,384. So the coefficient is −48,384.
The sign comes from (−3)³. Writing the general term with b = −3, not 3, keeps it automatically.
How do you find the term independent of x?
Example. Find the term independent of x in (x² − 2/x)⁹.
General term: ⁹Cᵣ (x²)⁹⁻ʳ (−2/x)ʳ = ⁹Cᵣ (−2)ʳ x¹⁸⁻²ʳ⁻ʳ = ⁹Cᵣ (−2)ʳ x¹⁸⁻³ʳ.
Independent of x means 18 − 3r = 0, so r = 6.
The term is ⁹C₆ × (−2)⁶ = 84 × 64 = 5376. It is the 7th term.
If solving for r gives a fraction or a number outside 0 to n, there is no such term. That is a valid answer.
Which terms are the middle terms?
If n is even, there is one middle term: T(n/2 + 1).
If n is odd, there are two: T((n + 1)/2) and T((n + 3)/2).
Example (n even). In (x + 1/x)¹⁰ the middle term is T₆, with r = 5: ¹⁰C₅ x⁵ x⁻⁵ = 252.
Example (n odd). In (2x − y)⁷ the middle terms are T₄ and T₅.
T₄ (r = 3): ⁷C₃ (2x)⁴ (−y)³ = 35 × 16 × (−1) x⁴y³ = −560x⁴y³.
T₅ (r = 4): ⁷C₄ (2x)³ (−y)⁴ = 35 × 8 x³y⁴ = 280x³y⁴.
How do you use properties of the coefficients?
Sum of all coefficients: put x = 1. For (2x − 3)⁸, that is (2 − 3)⁸ = 1.
Sum of binomial coefficients: ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ, from (1 + 1)ⁿ.
Alternating sum: ⁿC₀ − ⁿC₁ + ⁿC₂ − … = 0, from (1 − 1)ⁿ.
Remainders: 9⁵⁰ = (8 + 1)⁵⁰. Every term except the last contains a power of 8, so the remainder on dividing by 8 is 1.
Approximations: (1.01)¹⁰ = (1 + 0.01)¹⁰ ≈ 1 + 10(0.01) + 45(0.0001) + 120(0.000001) = 1 + 0.1 + 0.0045 + 0.00012 = 1.10462, to five decimal places.
Which mistakes cost the most marks?
Dropping the sign of b. Write (−3)ʳ, not 3ʳ, in the general term.
Confusing r with the term number. r = 6 is the 7th term.
Forgetting the power on a coefficient, writing 2x⁵ instead of (2x)⁵ = 32x⁵.
Mixing up "coefficient" and "binomial coefficient". In (2x − 3)⁸ the binomial coefficient of the x⁵ term is ⁸C₃ = 56; its coefficient is −48,384.
Taking one middle term when n is odd.
A practice routine
Worked examples help most at the start of a topic, but only if you then vary them. In five experiments on algebra, Sweller and Cooper (1985) found that students who studied worked examples solved similar problems faster and with fewer errors than students who solved conventional problems. The benefit was specific to problems with the same structure as the examples, so plan for variety too.
Study two worked examples per type (coefficient, independent term, middle term), then solve two of each yourself without looking.
Change the structure (weekly): products like (1 + x)³(1 − x)⁴, expressions with x⁻¹ or √x, and remainder questions.
Mix with counting: ⁿCᵣ comes from permutations and combinations, and some JEE questions combine the two.
Check one coefficient by expanding a small case, such as (2x − 3)³, fully.
When you set up the general term and still cannot get r, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can see where the power of x went wrong.
Tonight, find the coefficient of x⁴ in (3 − x/2)⁶, then check it by writing the full general term.
