Binomial Theorem

In this chapter: patterns in (a + b)ⁿ, Pascal's triangle, the binomial theorem for a positive integral index, the special cases (a − b)ⁿ and (1 + x)ⁿ, applications to numerical calculation, the general term, middle terms and the term independent of x.

Patterns in the expansions

Look at these closely and three things stand out. The expansion of (a + b)ⁿ has n + 1 terms. The power of a falls from n to 0 while the power of b rises from 0 to n. In every term the two powers add up to n.

Pascal's triangle

Write the coefficients in rows, one row for each power. Each row begins and ends with 1, and every other entry is the sum of the two entries just above it. This arrangement is called Pascal's triangle.

The entries of row n are exactly ⁿC₀, ⁿC₁, ..., ⁿC_n, and the addition rule is Pascal's identity ⁿC_r + ⁿC_r−1 = ⁿ⁺¹C_r from the previous chapter. The triangle is handy up to about n = 7; beyond that, use the formula.

The binomial theorem

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

The coefficients ⁿC_k are called binomial coefficients. Since ⁿC_k = ⁿC_n−k, coefficients equidistant from the two ends are equal.

Special cases

• (a − b)ⁿ: replace b by −b. The signs alternate: (x − y)ⁿ = ⁿC₀xⁿ − ⁿC₁xⁿ⁻¹y + ⁿC₂xⁿ⁻²y² − ... + (−1)^{n n}C_nyⁿ.
• (1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ... + ⁿC_nxⁿ.
• Putting x = 1: ⁿC₀ + ⁿC₁ + ... + ⁿC_n = 2ⁿ. So the sum of the coefficients of (1 + x)ⁿ is 2ⁿ.
• Putting x = −1: ⁿC₀ − ⁿC₁ + ⁿC₂ − ... = 0.

More generally, the sum of the coefficients of any polynomial in x is found by putting x = 1.

Worked example: Expand (2x − 3)⁴.
Solution: With a = 2x and b = −3: ⁴C₀(2x)⁴ + ⁴C₁(2x)³(−3) + ⁴C₂(2x)²(−3)² + ⁴C₃(2x)(−3)³ + ⁴C₄(−3)⁴ = 16x⁴ − 96x³ + 216x² − 216x + 81. Quick check: at x = 1 the left side is (−1)⁴ = 1, and 16 − 96 + 216 − 216 + 81 = 1.

Using the theorem for calculation

Split an awkward number into a round number plus or minus a small one. For (98)⁵, write 98 = 100 − 2:

The same idea compares numbers. Is (1.01)^1000000 bigger than 10000? Expanding, (1 + 0.01)^1000000 = 1 + 1000000 × 0.01 + (positive terms) = 1 + 10000 + (positive terms), which exceeds 10000. It also proves divisibility results: 6ⁿ = (1 + 5)ⁿ = 1 + 5n + 25 × (an integer), so 6ⁿ − 5n always leaves remainder 1 when divided by 25.

General term and middle terms

The general term formula is in the JEE Main syllabus, and most questions on this chapter use it. The (r + 1)th term of (a + b)ⁿ is

Note the shift: the term with ⁿC_r is the (r + 1)th term, so the 4th term uses r = 3.

• If n is even, there is one middle term, the (n/2 + 1)th term.
• If n is odd, there are two middle terms, the ((n + 1)/2)th and ((n + 3)/2)th terms.

So (x + y)¹⁰ has its middle term at T₆, while (x + y)⁷ has middle terms T₄ and T₅. The middle term, or the pair of middle terms, carries the greatest binomial coefficient.

Finding a particular term

Write T_r+1, collect all powers of x into a single power x^{(something in r)}, and set that exponent equal to what you want. For the term independent of x, set it equal to 0. The value of r must come out as a whole number from 0 to n; if it does not, that term does not exist.

Worked example: (a) Find the coefficient of x⁵ in (x + 3)⁸. (b) Find the term independent of x in (x² + 2/x)⁶.
Solution: (a) T_r+1 = ⁸C_r x^8−r 3^r. We need 8 − r = 5, so r = 3, and the coefficient is ⁸C₃ × 3³ = 56 × 27 = 1512.
(b) T_r+1 = ⁶C_r (x²)^6−r (2/x)^r = ⁶C_r 2^r x^12−3r. Setting 12 − 3r = 0 gives r = 4. The term is T₅ = ⁶C₄ × 2⁴ = 15 × 16 = 240.

Common mistakes: (1) Calling ⁿC_ra^n−rb^r the rth term; it is the (r + 1)th. (2) Losing the minus sign: in (a − b)ⁿ, take b as −b and raise the whole of it to the power. (3) Raising only the variable and forgetting the number, for example writing (2x)³ as 2x³. (4) Confusing the binomial coefficient ⁿC_r with the coefficient of the term, which also includes the numerical factors from a and b. (5) Accepting a fractional or negative r when looking for a term; such a term does not exist.

Exam focus

• Writing the general term and using it to find a given term, a coefficient, or the term independent of x.
• Middle terms for even and odd n.
• Sum of coefficients by substituting x = 1, and the identity ∑ ⁿC_r = 2ⁿ.
• Numerical uses: evaluating powers like (99)⁵, comparing large numbers, and remainder problems.
• Simplifying expressions like (a + b)ⁿ + (a − b)ⁿ, where alternate terms cancel.

Practice questions