MathsISC Class 11 · Binomial Theorem

Binomial Expansion and the General Term: Specific Terms and Coefficients

Updated 17 October 2026 · Fact-checked

For a positive integer n, (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ... + ⁿCₙbⁿ. It has n + 1 terms and the general term is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ. To find a particular term, a coefficient or the term independent of x, set the power of x in Tᵣ₊₁ equal to the required value and solve for r.

In every term the power of a falls as the power of b rises, and the two powers always add to n; the coefficient is nCr.

Illustration drawn for QuestionBuddy.

Understand Binomial Expansion and the General Term

Theorem. $(a + b)^n = \sum_{r=0}^{n} ^nC_r a^{n-r} b^r$ . Special forms: $(a - b)^{n}$ has alternating signs; $(1 + x)^{n} = \sum^{n} C_{r} x^{r}$ . The coefficients can be read from Pascal's triangle, in which each entry is the sum of the two above it. The powers of a fall from n to 0 while the powers of b rise from 0 to n, and the sum of the exponents in each term is n.

General term. $T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$ (the (r + 1)th term, r = 0, 1, ..., n). The k-th term uses r = k − 1. The term from the end: the pth term from the end of $(a + b)^{n}$ is the (n − p + 2)th from the beginning.

Finding a term. (1) Write $T_{r + 1}$ with all powers of x collected. (2) Equate the power of x to the required exponent to find r. (3) Substitute r. If r is not a non-negative integer (r ≤ n), the term does not exist.

Term independent of x. Set the power of x in $T_{r + 1}$ equal to 0.

Coefficient of a power. After finding r, the coefficient is the numerical part, including powers of the constants and signs.

Checking. Put x = 1 to check the sum of all coefficients, and use symmetry $^{n} C_{r} =^{n} C_{n - r}$ to reduce work.

Binomial Expansion and the General Term: formulas to remember

Binomial theorem: $(a + b)^n = \sum_{r=0}^{n} ^nC_r a^{n-r} b^r$; n + 1 terms.
General term: $T_{r+1} = ^nC_r a^{n-r} b^r$; r = 0, 1, ..., n.
Term from the end: $pt h f r o m t h ee n d = (n - p + 2) t h f r o m t h es t a r t$; Equivalent to expanding $(b + a)^{n}$ .
Coefficient sum: $P u t x = 1$; Sum of coefficients of $(1 + x)^{n}$ is $2^{n}$ .

Binomial Expansion and the General Term in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Compound growth approximation

Suppose an investment grows by 2% each year and the amount after 6 years is $(1.02)^{6}$ times the principal.

Use the binomial theorem to find the growth factor to four decimal places.

Try it first, then show the solutionSolution

$(1 + 0.02)^{6} = 1 + 6 (0.02) + 15 (0.02)^{2} + 20 (0.02)^{3} + ...$ .
$= 1 + 0.12 + 0.006 + 0.00016 + 0.0000024 + ...$ .
$\approx 1.1262$ .

Answer: About 1.1262.

Memory hook: Higher terms are very small.

Tricks to remember Binomial Expansion and the General Term

Short memory aids. Say each one aloud once, then use it on a question the same day.

r + 1 for the term number
The 5th term corresponds to r = 4.
Use it when: Index errors.
Power of x is zero
The term independent of x has zero total exponent of x.
Use it when: Constant term.

Binomial Expansion and the General Term: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Number of terms · 1 mark · MCQBoard level

The number of terms in the expansion of $(a + b)^{10}$ is:
(a) 10
(b) 11
(c) 12
(d) 9

Try it first, then show the solutionSolution

There are n + 1 = 11 terms.

Answer: (b) 11

Example 2: Expansion · 3 marks · Short answerBoard level

Expand $(2 x - 3)^{4}$ .

Try it first, then show the solutionSolution

$(2 x)^{4} +^{4} C_{1} (2 x)^{3} (- 3) +^{4} C_{2} (2 x)^{2} (- 3)^{2} +^{4} C_{3} (2 x) (- 3)^{3} + (- 3)^{4}$ . (1 mark)
$= 16 x^{4} - 4 \cdot 8 x^{3} \cdot 3 + 6 \cdot 4 x^{2} \cdot 9 - 4 \cdot 2 x \cdot 27 + 81$ . (1 mark)
$= 16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81$ . (1 mark)

Answer: 16x⁴ − 96x³ + 216x² − 216x + 81

Example 3: Term independent of x · 4 marks · Short answerHarder board level

Find the term independent of x in the expansion of $(2 x - \frac{1}{x ^{2}})^{9}$ .

Try it first, then show the solutionSolution

$T_{r + 1} =^{9} C_{r} (2 x)^{9 - r} (- \frac{1}{x ^{2}})^{r} =^{9} C_{r} 2^{9 - r} (- 1)^{r} x^{9 - 3 r}$ . (1.5 marks)
Independent of x: $9 - 3 r = 0$ , so $r = 3$ . (1 mark)
$T_{4} =^{9} C_{3} \cdot 2^{6} \cdot (- 1)^{3} = 84 \times 64 \times (- 1) = - 5376$ . (1.5 marks)

Answer: −5376

Example 4: Coefficient of a power · 4 marks · Short answerHarder board level

Find the coefficient of $x^{15}$ in $(x^{2} + \frac{1}{x})^{12}$ and the 5th term of $(1 - 2 x)^{7}$ .

Try it first, then show the solutionSolution

$T_{r + 1} =^{12} C_{r} x^{2 (12 - r)} x^{- r} =^{12} C_{r} x^{24 - 3 r}$ . (1 mark)
$24 - 3 r = 15$ gives $r = 3$ ; coefficient $=^{12} C_{3} = 220$ . (1 mark)
5th term of $(1 - 2 x)^{7}$ : r = 4: $^{7} C_{4} (- 2 x)^{4} = 35 \times 16 x^{4} = 560 x^{4}$ . (2 marks)

Answer: 220; 560x⁴.

Example 5: Terms in arithmetic progression · 5 marks · Long answerHOTS / competency

The coefficients of the 5th, 6th and 7th terms in the expansion of $(1 + x)^{n}$ are in AP. Find n and verify by computing the three coefficients for each value.

Try it first, then show the solutionSolution

The coefficients are $^{n} C_{4}$ , $^{n} C_{5}$ , $^{n} C_{6}$ ; AP gives $2 \cdot^{n} C_{5} =^{n} C_{4} +^{n} C_{6}$ . (1 mark)
Dividing by $^{n} C_{4}$ : $2 \cdot \frac{n - 4}{5} = 1 + \frac{( n - 4 ) ( n - 5 )}{30}$ . (1 mark)
Simplify: $n^{2} - 21 n + 98 = 0$ , so $n = 7$ or $n = 14$ . (1.5 marks)
n = 7: 35, 21, 7 ( $2 \times 21 = 42 = 35 + 7$ ). (0.75 mark)
n = 14: 1001, 2002, 3003 ( $2 \times 2002 = 4004 = 1001 + 3003$ ). (0.75 mark)

Answer: n = 7 or n = 14.

Where students lose marks in Binomial Expansion and the General Term

The slips examiners see most often, and what to write instead.

Mistake: Off-by-one in the term number
Using r = 5 for the 5th term.
Do this: The (r + 1)th term: for the 5th term r = 4.
Usually costs: 2 marks.
Mistake: Sign errors with negative terms
Dropping the minus sign for odd r.
Do this: Raise the whole term, including the sign, to the power r.
Usually costs: 1 to 2 marks.
Mistake: Mixing exponents of x
Combining 2(9 − r) incorrectly.
Do this: Collect the power of x carefully.
Usually costs: 1 mark.

Exam tips

Write the general term fully before simplifying.
State the equation for r clearly.
Check that r is a non-negative integer ≤ n.
Verify using a quick numerical check.

Practice questions on Binomial Expansion and the General Term

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The (r + 1)th term in the expansion of $(a + b)^{n}$ is:
(a) $^{n} C_{r} a^{r} b^{n - r}$
(b) $^{n} C_{r} a^{n - r} b^{r}$
(c) $^{n} C_{r + 1} a^{n - r} b^{r}$
(d) $^{n} C_{r} a^{n + r} b^{r}$
Hint
a falls from n.
Answer
(b) $^{n} C_{r} a^{n - r} b^{r}$
Q2 · 2 marks · Short answerBoard level
Expand $(1 + x)^{5}$ using Pascal's triangle.
Hint
Row 5: 1 5 10 10 5 1.
Answer
$1 + 5 x + 10 x^{2} + 10 x^{3} + 5 x^{4} + x^{5}$ .
Q3 · 3 marks · Short answerHarder board level
Find the 4th term of $(3 x - \frac{y}{2})^{6}$ .
Hint
r = 3.
Answer
$T_{4} =^{6} C_{3} (3 x)^{3} (- \frac{y}{2})^{3} = 20 \times 27 x^{3} \times (- \frac{y ^{3}}{8}) = - \frac{135}{2} x^{3} y^{3}$ .
Q4 · 3 marks · Short answerHarder board level
Find the coefficient of $x^{5}$ in the expansion of $(x + \frac{3}{x ^{2}})^{8}$ .
Hint
Powers: x^{8−3r}.
Answer
$T_{r + 1} =^{8} C_{r} x^{8 - r} 3^{r} x^{- 2 r} =^{8} C_{r} 3^{r} x^{8 - 3 r}$ ; $8 - 3 r = 5$ gives $r = 1$ : coefficient $8 \times 3 = 24$ .
Q5 · 3 marks · Short answerHarder board level
Using the binomial theorem, evaluate $(99)^{4}$ as $(100 - 1)^{4}$ .
Hint
Expand with alternating signs.
Answer
$1 0^{8} - 4 \times 1 0^{6} + 6 \times 1 0^{4} - 4 \times 1 0^{2} + 1 = 100000000 - 4000000 + 60000 - 400 + 1 = 96059601$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a number puzzle asks for the middle behaviour of $(1 + x)^{10}$ .
(i) How many terms are there? (1 mark)
(ii) Find the coefficient of $x^{3}$ . (1 mark)
(iii) Find the sum of the coefficients. (1 mark)
(iv) Find the greatest coefficient. (1 mark)
Hint
Coefficients are ¹⁰Cᵣ.
Answer
(i) 11. (ii) $^{10} C_{3} = 120$ . (iii) $2^{10} = 1024$ . (iv) $^{10} C_{5} = 252$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Number of terms?Show answer
n + 1.
2.General term?Show answer
ⁿCᵣ aⁿ⁻ʳ bʳ.
3.Sum of coefficients of (1+x)ⁿ?Show answer
2ⁿ.
4.Term independent of x?Show answer
Power of x equals zero.

Binomial Expansion and the General Term: frequently asked questions

What is Pascal's triangle?

A triangular array in which each number is the sum of the two above; row n gives the coefficients of (a + b)^n.

Why do alternate signs appear in $(a - b)^n$?

Because b^r becomes (-b)^r, which is negative for odd r.

Can the binomial theorem be used for negative or fractional n?

In this chapter n is a positive integer; for other values the expansion is an infinite series and requires |x| < 1.