Binomial Theorem JEE Notes PDF, Formulas, Practice Questions

Expanding (a + b)² by hand is easy enough, but (a + b)¹⁰ or (a + b)⁵⁰ is another matter. The binomial theorem gives a formula that writes out the expansion for any positive integer n without repeated multiplication, and JEE rarely asks for the whole expansion anyway. Practising JEE questions from this chapter helps students quickly identify the exact term, coefficient or power a problem asks for. These Binomial Theorem JEE notes cover the general term, the middle and greatest terms, coefficient identities, the multinomial theorem and binomial approximation, in a format built for fast revision.

Binomial Theorem JEE Notes: Important Concepts

Binomial: an algebraic expression with exactly two terms, such as (a + b), (x − 3) or (2x + 5y).

Binomial coefficient: the number ⁿCᵣ = n!/[r!(n − r)!], read "n choose r", which counts the ways of choosing r objects from n. It is also written C(n, r).

Factorial: n! means n × (n − 1) × (n − 2) × ... × 2 × 1, and by convention 0! = 1.

Worked example: compute ⁵C₂.

⁵C₂ = 5!/(2!·3!) = (5 × 4 × 3!)/(2 × 1 × 3!) = 20/2 = 10

Cancelling the larger factorial rather than expanding everything is the habit worth building here, since JEE coefficients often involve numbers far too large to write out.

The Binomial Theorem and the General Term

Statement of the Theorem

For any positive integer n and any real numbers a and b:

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

Written out in full:

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... + ⁿCₙbⁿ

Four things to note: n is a positive integer, a and b can be any real or complex numbers, r runs from 0 to n, and the expansion has exactly n + 1 terms.

Worked example: expand (x + y)⁴.

(x + y)⁴ = ⁴C₀x⁴ + ⁴C₁x³y + ⁴C₂x²y² + ⁴C₃xy³ + ⁴C₄y⁴ = 1·x⁴ + 4x³y + 6x²y² + 4xy³ + 1·y⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

Worked example: expand (2x − 3)³.

Here a = 2x, b = −3, n = 3. (2x − 3)³ = ³C₀(2x)³ + ³C₁(2x)²(−3) + ³C₂(2x)(−3)² + ³C₃(−3)³ = 8x³ + 3(4x²)(−3) + 3(2x)(9) + (−27) = 8x³ − 36x² + 54x − 27

JEE tip: when b is negative, as in (a − b)ⁿ, the signs alternate: positive, negative, positive, negative. The sign of Tᵣ₊₁ is (−1)ʳ.

General Term

You rarely need the whole expansion. A JEE question typically asks for the 5th term, or the term containing x⁷, and the general term formula jumps straight there.

In the expansion of $$(a+b)^n$$, the general term is $$T_{r+1}=\binom{n}{r}a^{n-r}b^r$$.

So r = 0 gives T₁, r = 1 gives T₂, and r = n gives Tₙ₊₁, the last term.

Note: watch the indexing. Tᵣ₊₁ corresponds to r, not to r + 1, so to find the k-th term set r = k − 1. Off-by-one here is the single most common error in the chapter.

Worked example: find the 6th term of (2x + 3y)⁸.

The 6th term is T₆ = T₅₊₁, so r = 5. T₆ = ⁸C₅(2x)⁸⁻⁵(3y)⁵ = ⁸C₅(2x)³(3y)⁵ = 56 × 8x³ × 243y⁵ = 56 × 8 × 243 × x³y⁵ = 108864 x³y⁵

Worked example: find the term containing x⁵ in the expansion of (x² + 1/x)⁷.

Here a = x², b = 1/x = x⁻¹, n = 7. Tᵣ₊₁ = ⁷Cᵣ(x²)⁷⁻ʳ(x⁻¹)ʳ = ⁷Cᵣ x¹⁴⁻²ʳ · x⁻ʳ = ⁷Cᵣ x¹⁴⁻³ʳ For the term containing x⁵: 14 − 3r = 5, so 3r = 9 and r = 3. T₄ = ⁷C₃x⁵ = 35x⁵

Worked example: find the term independent of x in the expansion of (x − 2/x²)⁹.

Here a = x, b = −2/x² = −2x⁻², n = 9. Tᵣ₊₁ = ⁹Cᵣ(x)⁹⁻ʳ(−2x⁻²)ʳ = ⁹Cᵣ(−2)ʳ x⁹⁻ʳ⁻²ʳ = ⁹Cᵣ(−2)ʳ x⁹⁻³ʳ For the term independent of x, the power must be zero: 9 − 3r = 0, so r = 3. T₄ = ⁹C₃(−2)³ = 84 × (−8) = −672

The method is the same every time. Write the general term, collect the power of x into a single exponent, set that exponent to whatever the question demands, and solve for r.

Middle Term, Greatest Coefficient and Greatest Term

Middle Term

Which term sits in the middle depends on the parity of n, since the expansion has n + 1 terms.

Worked example: find the middle term of (x + 2)⁶.

n = 6 is even, so the middle term has index 6/2 + 1 = 4, meaning T₄ with r = 3. T₄ = ⁶C₃x³ · 2³ = 20 × x³ × 8 = 160x³

Worked example: find the middle terms of (1 + x)⁷.

n = 7 is odd, so the two middle terms have indices (7 + 1)/2 = 4 and (7 + 3)/2 = 5. T₄ = ⁷C₃x³ = 35x³ T₅ = ⁷C₄x⁴ = 35x⁴

Greatest Binomial Coefficient

Among ⁿC₀, ⁿC₁, ..., ⁿCₙ, the largest sits in the middle of the row.

Worked example: find the greatest binomial coefficient in (1 + x)¹⁰.

n = 10 is even, so the greatest coefficient is ¹⁰C₅ = 252.

Numerically Greatest Term

This is a different question from the greatest coefficient, since it accounts for the actual values of a and b. Here you want the term with the largest absolute value, found by tracking where the ratio of consecutive terms crosses 1.

$$\left|\frac{T_{r+1}}{T_r}\right|=\frac{n-r+1}{r}\left|\frac{b}{a}\right|$$

Set that ratio greater than or equal to 1 and solve for r. The greatest term corresponds to the largest integer r satisfying the inequality, which is where the ratio last stays at or above 1 before dropping below it.

Worked example: find the numerically greatest term in (1 + 3)⁶, that is with a = 1, b = 3, n = 6.

|Tᵣ₊₁/Tᵣ| = [(6 − r + 1)/r] · 3 = 3(7 − r)/r ≥ 1 21 − 3r ≥ r, so 21 ≥ 4r and r ≤ 5.25 The terms keep increasing for r = 1, 2, 3, 4, 5, so the largest is T₆ at r = 5. T₆ = ⁶C₅(1)¹(3)⁵ = 6 × 243 = 1458 Check: T₇ = ⁶C₆(3)⁶ = 729, and indeed T₆ > T₇, confirming T₆ is greatest.

Properties and Identities of Binomial Coefficients

These identities are tested heavily, especially in summation problems.

Basic Properties

JEE tip: symmetry saves real time. ¹⁰⁰C₉₈ = ¹⁰⁰C₂ = 4950, which is a two-second calculation instead of a nightmare.

Worked example: verify Pascal's identity for n = 5, r = 2.

⁵C₂ + ⁵C₁ = 10 + 5 = 15 ⁶C₂ = 6!/(2!·4!) = 15 ✓

Sum of Binomial Coefficients

Each of these comes from substituting particular values of a and b into the expansion, which is a powerful trick in its own right.

Worked example: find ⁸C₀ + ⁸C₁ + ⁸C₂ + ... + ⁸C₈.

By the sum identity, Σ(r = 0 to 8) ⁸Cᵣ = 2⁸ = 256

Worked example: find ¹⁰C₀ + ¹⁰C₂ + ¹⁰C₄ + ... + ¹⁰C₁₀.

This is the sum of the even-indexed coefficients, so it equals 2¹⁰⁻¹ = 2⁹ = 512

JEE tip: to find the sum of the coefficients of any polynomial f(x), substitute x = 1. The sum of the coefficients of (3x − 2)⁵ is f(1) = (3 − 2)⁵ = 1.

Other Important Identities

Vandermonde's identity splits a "choose" across two groups. The weighted sum follows from differentiating (1 + x)ⁿ and setting x = 1, which is worth remembering as a technique rather than as one more formula.

Worked example: find Σ(r = 0 to 6) r · ⁶Cᵣ.

Using the identity, Σ r·⁶Cᵣ = 6 · 2⁶⁻¹ = 6 × 32 = 192

Pascal's Triangle

Each entry is the sum of the two directly above it, which is exactly Pascal's identity in visual form. It is the quickest way to get coefficients for small n.

Multinomial Theorem and Applications to Approximation

Multinomial Theorem

The binomial theorem handles two terms, and the multinomial theorem generalises it to three or more. The full theorem is rarely asked directly in JEE Mains, but the general term is useful.

(x₁ + x₂ + ... + xₖ)ⁿ = Σ [n!/(r₁! r₂! ... rₖ!)] x₁^r₁ x₂^r₂ ... xₖ^rₖ

The sum runs over all non-negative integers r₁, r₂, ..., rₖ with r₁ + r₂ + ... + rₖ = n. The number of terms in the expansion is ⁿ⁺ᵏ⁻¹C(k−1), and the quantity n!/(r₁!r₂!...rₖ!) is called a multinomial coefficient.

Worked example: find the coefficient of x²y²z in (x + y + z)⁵.

Here r₁ = 2, r₂ = 2, r₃ = 1, and r₁ + r₂ + r₃ = 5 = n ✓ Coefficient = 5!/(2!·2!·1!) = 120/(4 × 1) = 30

Binomial Approximation

When x is small, the first few terms of (1 + x)ⁿ carry almost all the value and the rest can be dropped.

When $$|x|\ll1$$: $$(1+x)^n\approx1+nx+\frac{n(n-1)}{2}x^2+\cdots$$

For a rough first-order estimate, (1 + x)ⁿ ≈ 1 + nx.

Worked example: approximate (1.01)¹⁰.

Write 1.01 = 1 + 0.01, so x = 0.01 and n = 10. (1.01)¹⁰ ≈ 1 + 10(0.01) + (10 × 9/2)(0.01)² = 1 + 0.1 + 45 × 0.0001 = 1 + 0.1 + 0.0045 = 1.1045 The actual value is 1.10462..., so two terms of correction already get very close.

Remainder Problems

Worked example: find the remainder when 7¹⁰³ is divided by 25.

7¹⁰³ = 7 · 7¹⁰² = 7 · (7²)⁵¹ = 7(49)⁵¹ = 7(50 − 1)⁵¹ Expanding (50 − 1)⁵¹ = Σ(r = 0 to 51) ⁵¹Cᵣ 50⁵¹⁻ʳ(−1)ʳ, every term carrying 50ᵏ with k ≥ 1 is divisible by 25. The only survivor is ⁵¹C₅₁(−1)⁵¹ = −1. So (50 − 1)⁵¹ = 25m − 1 for some integer m. 7¹⁰³ = 7(25m − 1) = 175m − 7 = 25(7m) − 7 = 25(7m − 1) + 25 − 7 = 25(7m − 1) + 18 Remainder = 18

JEE tip: for remainder questions, rewrite the base as a multiple of the divisor plus or minus a small number, then expand and keep only the last term. Everything else is divisible by the divisor and disappears.

Binomial Theorem Formula Sheet and JEE Important Points

The formulas below cover the core identities used in this chapter. During revision, students can also use the JEE formula sheet to revise these results along with formulas from other JEE Mathematics chapters.

Formulas at a Glance

Common Mistakes to Avoid

1. Confusing Tᵣ₊₁ with the r-th term. For the k-th term, use r = k − 1, so the 6th term needs r = 5.
2. Dropping the sign inside b. In (2x − 3)³ the second term is b = −3, and the negative sign must ride along through every power.
3. Forgetting to raise the whole coefficient to the power. In (2x)³ the 2 is cubed too, giving 8x³ rather than 2x³.
4. Mishandling negative exponents. With b = 1/x = x⁻¹, the powers subtract, so collect everything into one exponent before solving.
5. Confusing the greatest coefficient with the greatest term. The first depends only on n, the second on the actual values of a and b.
6. Giving one middle term when n is odd. An odd n means n + 1 terms, an even count, so there are two middle terms.
7. Assuming the expansion always has n terms. It has n + 1.
8. Reaching for the wrong sum identity. The even-indexed and odd-indexed sums are each 2ⁿ⁻¹, not 2ⁿ.
9. Using the approximation when x is not small. The expansion (1 + x)ⁿ ≈ 1 + nx only holds when the magnitude of x is well under 1.
10. Keeping too many terms in remainder problems. Only the final term survives, since every other one carries the divisor as a factor.

Quick Revision Notes for the Binomial Theorem

• (a + b)ⁿ has n + 1 terms, and powers of a fall while powers of b rise, with the two exponents always summing to n.
• General term Tᵣ₊₁ = ⁿCᵣaⁿ⁻ʳbʳ. For a specific term, set r = k − 1. For a specific power of x, collect the exponent and solve for r.
• Term independent of x means the collected exponent equals zero.
• Middle term: one when n is even, at index (n/2) + 1, and two when n is odd, at (n + 1)/2 and (n + 3)/2.
• Greatest coefficient sits at the middle of the row, ⁿC(n/2) for even n.
• Numerically greatest term comes from the consecutive ratio [(n − r + 1)/r]·|b/a| ≥ 1.
• ⁿCᵣ = ⁿC(n−r) and ⁿCᵣ + ⁿC(r−1) = ⁿ⁺¹Cᵣ, which is why Pascal's triangle works.
• Substituting a = b = 1 gives 2ⁿ, and a = 1 with b = −1 gives 0. Substituting x = 1 gives the sum of a polynomial's coefficients.
• Σ r·ⁿCᵣ = n·2ⁿ⁻¹ and Σ(ⁿCᵣ)² = ²ⁿCₙ.
• For small x, (1 + x)ⁿ ≈ 1 + nx + [n(n − 1)/2]x².
• For remainders, write the base as (multiple of the divisor ± small number) and expand.