Binomial Theorem

🎯 Key Points

• General term T(r+1) = nCr aⁿ⁻ʳ bʳ — to find a SPECIFIC term (e.g. "6th term"), set r = (term number)-1
• "Term independent of x" or "constant term": set the net exponent of x in T(r+1) to ZERO and solve for r, then substitute back
• Sum of ALL coefficients (put a=b=1): 2ⁿ; Sum of coefficients with alternating sign (put a=1,b=-1): 0; Sum of even-indexed coefficients: 2ⁿ⁻¹
• Greatest binomial COEFFICIENT (not greatest term value) is always the middle term, since nCr peaks at r=n/2

Binomial Theorem

The Binomial Theorem gives the expansion of (a + b)ⁿ for any positive integer n without actual multiplication.

Binomial Expansion

(a + b)ⁿ = nC0 aⁿ + nC1 aⁿ⁻¹b + nC2 aⁿ⁻²b² + ... + nCn bⁿ = Σ nCr aⁿ⁻ʳ bʳ (r = 0 to n)

General Term

T(r+1) = nCr × aⁿ⁻ʳ × bʳ (the (r+1)th term of the expansion)

Middle Term(s)

• If n is even: one middle term: T(n/2 + 1)
• If n is odd: two middle terms: T((n+1)/2) and T((n+3)/2)

Binomial Coefficients: Key Identities

• nC0 + nC1 + nC2 + ... + nCn = 2ⁿ (put a=b=1)
• nC0 − nC1 + nC2 − ... = 0 (put a=1, b=−1)
• nC0 + nC2 + nC4 + ... = 2ⁿ⁻¹ (sum of even-indexed coefficients)
• nCr = nC(n−r) (symmetry)
• Pascal identity: nCr = (n−1)C(r−1) + (n−1)Cr

Greatest Term

Find r such that T(r+1)/T(r) ≥ 1 and T(r+2)/T(r+1) ≤ 1. The value of r where ratio crosses 1 gives the numerically greatest term.

Special Cases

• (1+x)ⁿ = 1 + nx + n(n-1)/2! x² + ... (important for finding specific terms)
• Coefficient of xʳ in (1+x)ⁿ is nCr

Finding a Specific Term

To find a particular term (such as the 6th term, or the term containing x^9), use the general term formula T(r+1) = nCr a^(n-r) b^r and match the position or the power required. For the "k-th term", set r = k - 1.

Worked example: Find the 5th term in the expansion of (x/2 + 2/x)^8.

Here n = 8, a = x/2, b = 2/x. The 5th term means r + 1 = 5, so r = 4.

T5 = 8C4 (x/2)^4 (2/x)^4 = 70 x (x^4/16) x (16/x^4) = 70.

Notice how the x-powers cancel exactly, this is a common JEE trick: when a term is asked to be "independent of x" or a "constant term", set the net power of x in T(r+1) to zero and solve for r.

Term Independent of x

Worked example: Find the term independent of x in (2x^2 - 1/x)^6.

T(r+1) = 6Cr (2x^2)^(6-r) (-1/x)^r = 6Cr 2^(6-r) (-1)^r x^(12-2r-r) = 6Cr 2^(6-r) (-1)^r x^(12-3r).

For the term independent of x, set 12 - 3r = 0, giving r = 4. So the term is 6C4 x 2^2 x (-1)^4 = 15 x 4 x 1 = 60.

Pascal's Triangle and Coefficient Patterns

• Each row of Pascal's triangle lists nCr for r = 0 to n, and every entry equals the sum of the two entries above it: nCr = (n-1)C(r-1) + (n-1)Cr.
• The coefficients of (a+b)^n are symmetric: nC0, nC1, ..., nCn read the same forwards and backwards (since nCr = nC(n-r)).
• For (1+x)^n, the ratio of consecutive coefficients is nCr / nC(r-1) = (n-r+1)/r, useful for quickly building up a row without recomputing factorials.

Common JEE Trick Points

• When the binomial is (a - b)^n, the signs of terms alternate: T(r+1) = nCr a^(n-r) (-b)^r, so even r gives a positive term and odd r gives a negative term.
• If asked for the term with the greatest binomial coefficient (not greatest numerical value), it is simply the middle term, since nCr is maximum at r = n/2 (or the two middle values when n is odd).
• For expansions like (1+x)^n, putting x = 1 gives the sum of all coefficients; this is a quick way to check a multi-part question without expanding fully.

More Properties of Binomial Coefficients

Writing C0, C1, ..., Cn for nC0, nC1, ..., nCn:

• C0 + C1 + C2 + ... + Cn = 2ⁿ (put x = 1 in (1+x)ⁿ).
• C1 + 2C2 + 3C3 + ... + nCn = n·2ⁿ⁻¹ (differentiate (1+x)ⁿ, then put x = 1).
• C0 + C1/2 + C2/3 + ... + Cn/(n+1) = (2ⁿ⁺¹ − 1)/(n+1) (integrate (1+x)ⁿ from 0 to 1).
• C0² + C1² + C2² + ... + Cn² = (2n)Cn (coefficient of xⁿ in (1+x)ⁿ·(1+x)ⁿ).
• Differentiating or integrating the standard expansion is the routine trick for evaluating such coefficient sums.

Greatest Binomial Coefficient

The greatest coefficient in (1+x)ⁿ is the middle binomial coefficient, since nCr rises up to r = n/2 and then falls. For even n the single greatest coefficient is nC(n/2); for odd n there are two equal greatest coefficients, nC((n−1)/2) and nC((n+1)/2). This is distinct from the numerically greatest TERM, whose position also depends on the actual values of a and x.

Applications: Divisibility and Remainders

Writing a number as (multiple ± small remainder) and expanding with the binomial theorem isolates the remainder.

• Example: 8ⁿ = (1 + 7)ⁿ = 1 + 7n + 7²(...) , so every term after the first is divisible by 7 — hence 8ⁿ leaves remainder 1 when divided by 7.
• For last digit(s), work modulo 10 or 100: since 3⁴ = 81 ends in 1, the last digit of powers of 3 repeats with period 4.
• General strategy: express the base as (k·m ± 1) so that all but one or two binomial terms vanish under the required modulus.

Applications: Approximation for Small x

When |x| is small, higher powers x², x³, ... become negligible, so (1 + x)ⁿ ≈ 1 + nx to first order. Keeping one more term gives (1 + x)ⁿ ≈ 1 + nx + n(n−1)/2 · x². This estimates values such as (1.02)⁵ ≈ 1 + 5(0.02) = 1.10 quickly without full expansion, and underlies error-estimation in physics and engineering.

🚀 JEE Advanced Edge

Finding the numerically greatest term precisely: Unlike the greatest COEFFICIENT (always the middle term), the greatest numerical TERM in (a+x)ⁿ depends on the actual values of a and x — compute the ratio T(r+1)/T(r) = [(n-r+1)/r]·(x/a), set it ≥1 to find the range of r where terms are still increasing, and the term just before the ratio drops below 1 is the greatest term. This is a genuinely different (and more involved) calculation than the middle-term shortcut.

Multinomial theorem as a generalisation: (a+b+c)ⁿ expands with general term n!/(p!q!r!) · aᵖbᵠcʳ where p+q+r=n — this extends the binomial coefficient nCr (which is really just the 2-variable case n!/(r!(n-r)!)) to any number of terms, and is the standard tool for "coefficient of a specific term in a trinomial expansion" problems.

Worked problem: Find the coefficient of x⁵ in the expansion of (1+x)⁷(1-x)⁵. Approach: Write (1+x)⁷(1-x)⁵ = (1+x)²·(1-x²)⁵ since (1+x)⁵(1-x)⁵=(1-x²)⁵. Expand (1-x²)⁵ = Σ5Ck(-1)kx²k and (1+x)² = 1+2x+x². The x⁵ term needs odd total power, only possible from x²·x²k terms combined with the x¹ part of (1+x)²: coefficient of x⁵ = 2×[coefficient of x⁴ in (1-x²)⁵] = 2×5C2×(-1)² = 2×10 = 20.

Worked Example: General Term of a Binomial Expansion

Find the term containing x³ in the expansion of (2x − 1/x)⁷.

General term: T(r+1) = C(7,r) · (2x)^(7−r) · (−1/x)^r = C(7,r) · 2^(7−r) · (−1)^r · x^(7−r−r) = C(7,r) · 2^(7−r) · (−1)^r · x^(7−2r).

For x³: 7 − 2r = 3 → r = 2. T₃ = C(7,2) · 2⁵ · (−1)² · x³ = 21 · 32 · 1 · x³ = 672x³. Always write the general term first and set the power equal to the required value to find r.

Worked Example: Middle Term

Find the middle term of (x + 1/x)⁸.

n = 8 (even), so there is one middle term: T(n/2 + 1) = T₅. T₅ = C(8,4) · x⁴ · (1/x)⁴ = 70 · x⁴ · x⁻⁴ = 70. For even n, middle term is T(n/2+1); for odd n, there are two middle terms T((n+1)/2) and T((n+3)/2).