The Binomial Theorem and the General Term: Find Any Term Without Expanding

You can expand (a + b)² in your head, and (a + b)³ with a little effort. But what is the coefficient of x⁶ in (x² + 1/x)⁹? Nobody expands nine brackets in an exam hall. The binomial theorem, and one formula for its general term, lets you jump straight to the single term you need. This topic is important for JEE Main and JEE Advanced, and it is a reliable source of marks every year.

Quick Summary

• (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCₙbⁿ, for any positive integer n.

• The expansion has n + 1 terms, and the powers of a and b in every term add up to n.

• General term: T(r+1) = ⁿCᵣ · aⁿ⁻ʳ · bʳ. Note that r+1 is the position, and r is the power of the second term.

• To find a particular term, write the power of x in T(r+1), set it equal to what you want, and solve for r.

• r must be a whole number from 0 to n; if it is not, that term does not exist.

Where the formula comes from

Start with something you know: (a + b)² = a² + 2ab + b². Multiply by (a + b) again and you get a³ + 3a²b + 3ab² + b³. The coefficients 1, 2, 1 and 1, 3, 3, 1 are rows of Pascal's triangle. But why do these numbers appear?

Write (a + b)ⁿ as n identical brackets multiplied together: (a + b)(a + b)…(a + b). To build one term of the expansion, you pick either a or b from each bracket. Suppose you pick b from exactly r of the brackets. Then you must pick a from the other n − r brackets, so the term looks like aⁿ⁻ʳbʳ. How many different ways can you choose which r brackets give b? That is exactly ⁿCᵣ = n! / (r! (n − r)!), the number of ways to choose r objects from n. Every such choice gives the same product aⁿ⁻ʳbʳ, so they add up to ⁿCᵣ aⁿ⁻ʳbʳ.

That is the whole theorem: the coefficient is simply a count of choices. Once you see it this way, you will not need to memorise the formula blindly. Here ⁿCᵣ means "n choose r", and r takes the values 0, 1, 2, …, n.

The general term

The term containing bʳ is the (r + 1)th term, because r starts at 0:

T(r+1) = ⁿCᵣ · aⁿ⁻ʳ · bʳ

Here T(r+1) means the (r + 1)th term, a is the first quantity inside the bracket and b is the second. Always take a as the first term and b as the second exactly as written in the bracket, including any minus sign. So for (x − 2/x²)¹², you have a = x and b = −2/x², and the sign goes with b.

A useful consequence: the coefficients are symmetric, because ⁿCᵣ = ⁿCₙ₋ᵣ. The kth term from the beginning and the kth term from the end have the same binomial coefficient.

Worked example 1: a specific power of x

Question: Find the coefficient of x⁶ in the expansion of (x² + 1/x)⁹.

Step 1: Identify a = x², b = 1/x = x⁻¹, n = 9.

Step 2: Write the general term. T(r+1) = ⁹Cᵣ (x²)⁹⁻ʳ (x⁻¹)ʳ.

Step 3: Simplify the power of x. (x²)⁹⁻ʳ = x¹⁸⁻²ʳ, and (x⁻¹)ʳ = x⁻ʳ. Multiplying, the power is 18 − 2r − r = 18 − 3r.

Step 4: We want x⁶, so 18 − 3r = 6, which gives 3r = 12 and r = 4. This is a whole number between 0 and 9, so the term exists.

Step 5: The coefficient is ⁹C₄ = 9! / (4! × 5!) = (9 × 8 × 7 × 6) / (4 × 3 × 2 × 1) = 3024 / 24 = 126.

Answer: 126. The same setup tells you the term independent of x: put 18 − 3r = 0, so r = 6, and the term is ⁹C₆ = 84.

Worked example 2: a term with a sign and a constant

Question: Find the term independent of x in (x − 2/x²)¹².

Step 1: a = x, b = −2/x² = −2x⁻², n = 12.

Step 2: T(r+1) = ¹²Cᵣ · x¹²⁻ʳ · (−2)ʳ · x⁻²ʳ.

Step 3: The power of x is 12 − r − 2r = 12 − 3r. For a term independent of x, set 12 − 3r = 0, so r = 4.

Step 4: The term is ¹²C₄ × (−2)⁴ = 495 × 16. Here ¹²C₄ = (12 × 11 × 10 × 9) / 24 = 11880 / 24 = 495.

Step 5: 495 × 16 = 7920. Answer: 7920, and it is the 5th term, since r + 1 = 5.

Mistakes students actually make

• Confusing r with the term number. If the question asks for the 7th term, then r = 6, not 7. Students who put r = 7 get the wrong term every time.

• Dropping the sign. In (x − 2/x²), the "−2" belongs to b, and (−2)ʳ is positive for even r and negative for odd r. Forgetting this flips the answer's sign.

• Forgetting to raise the constant to the power. In (2x + 3)⁵ the term is ⁵Cᵣ (2x)⁵⁻ʳ 3ʳ, so you must compute 2⁵⁻ʳ and 3ʳ as well. The coefficient is not just ⁵Cᵣ.

• Mixing up the order of a and b. Swapping them changes which power r applies to, and your equation for r will be wrong.

• Accepting a non-integer r. If solving gives r = 2.5 or r = −1, the required term does not exist and the answer is 0 (or "no such term"). Do not round.

How it is asked in JEE

• "Find the coefficient of xᵏ in (…)ⁿ", the most common form, solved exactly as in Example 1.

• "Find the term independent of x", where you set the total power to 0.

• "Find the (r + 1)th term from the end", where you use the symmetry of the expansion: it is the (n − r + 1)th term from the beginning.

• "If the coefficients of two consecutive or given terms are in a ratio or in A.P.", where you set up an equation in n and r.

• Questions on the sum of coefficients, where you put x = 1 in the expansion.

Pro Tip: Never expand to look for a term. Combine the powers of x in T(r+1) first into a single expression like 18 − 3r. Then set it equal to the target power and solve for r. One linear equation replaces pages of algebra, and it also immediately tells you whether the term exists. If you keep a rough practice habit of checking that 0 ≤ r ≤ n at the end, you will avoid most careless marks lost in this chapter. A quick sanity check is that the coefficients of all terms add up to the value of the whole expression at x = 1, which for (x² + 1/x)⁹ is 2⁹ = 512.

Frequently Asked Questions

How many terms are there in the expansion of (a + b)ⁿ?

There are n + 1 terms, because r runs from 0 to n. For example, (a + b)⁵ has 6 terms. This is valid when n is a positive integer; for other powers the theorem gives an infinite series.

Why is the general term called T(r+1) and not T(r)?

Because the first term has r = 0, the second has r = 1, and so on. So the term with bʳ is the (r + 1)th term. This small offset is the source of many errors, so always write down which term number the question asks for and subtract 1 to get r.

What is the difference between a coefficient and a term?

A term includes the variable part, for example 126x⁶. The coefficient is only the numerical factor, 126. Read the question carefully, since the coefficient of x⁶ and the term containing x⁶ are different answers.

What if the expression has a negative sign, like (a − b)ⁿ?

Treat it as (a + (−b))ⁿ. The general term is then ⁿCᵣ aⁿ⁻ʳ (−b)ʳ, so terms alternate in sign, starting positive. Keep the minus sign with b, never with a.

How do I find the sum of all the coefficients?

Put the variable equal to 1 (and in the case of the constant term as well, nothing else changes). For (1 + x)ⁿ this gives 2ⁿ. For (x² + 1/x)⁹ it gives (1 + 1)⁹ = 512. Putting x = −1 gives the alternating sum instead.

Can r be a fraction or negative?

No. r counts brackets from which b is chosen, so it must be a whole number from 0 to n. If your equation gives anything else, the required term does not exist in the expansion.

Stuck on a binomial problem or a step in the examples above? Drop your doubt in the comments, and share this with a fellow JEE aspirant who is still expanding brackets by hand.