Binomial Theorem ISC Class 11: Pascal's Triangle, General and Middle Terms

What Binomial Theorem is about

Weightage: Binomial Theorem belongs to the unit Algebra, which CISCE weights at 26 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Pascal's triangle and its significance

Write 1 at the top; each row starts and ends with 1, and every inner entry is the sum of the two entries just above it. Row n (counting the top row as row 0) gives the coefficients of (x + y)ⁿ: ⁿC₀, ⁿC₁, …, ⁿCₙ. The building rule is Pascal's rule ⁿCᵣ₋₁ + ⁿCᵣ = ⁿ⁺¹Cᵣ. Significance: each row is symmetric (ⁿCᵣ = ⁿCₙ₋ᵣ), the entries of row n add to 2ⁿ, and the numbers count selections, so the triangle links algebra with counting. For small n it is the fastest way to expand.

Worked example

Use Pascal's triangle to expand (2x + 3)⁴.

Try it first, then show the solutionSolution

1. Row 4: 1, 4, 6, 4, 1
2. (2x)⁴ + 4(2x)³(3) + 6(2x)²(3²) + 4(2x)(3³) + 3⁴
3. = 16x⁴ + 96x³ + 216x² + 216x + 81

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

Statement and proof of the binomial theorem

Statement: for any positive integer n, (x + y)ⁿ = ⁿC₀xⁿ + ⁿC₁xⁿ⁻¹y + ⁿC₂xⁿ⁻²y² + … + ⁿCₙyⁿ. Proof (by induction): for n = 1 both sides are x + y. Assume it for n = k. Then (x + y)ᵏ⁺¹ = (x + y)(x + y)ᵏ; multiplying the assumed expansion by x and by y and collecting the terms containing xᵏ⁺¹⁻ʳyʳ gives coefficient ᵏCᵣ + ᵏCᵣ₋₁ = ᵏ⁺¹Cᵣ by Pascal's rule, while the first and last coefficients are 1 = ᵏ⁺¹C₀ = ᵏ⁺¹Cₖ₊₁. So the result holds for k + 1, and hence for all n. Counting view: in the product of n brackets (x + y), a term xⁿ⁻ʳyʳ arises by choosing y from r brackets, which can be done in ⁿCᵣ ways.

Worked example

Use the theorem to find (√3 + √2)⁴ + (√3 − √2)⁴.

Try it first, then show the solutionSolution

1. In the sum, odd powers of √2 cancel: 2[(√3)⁴ + ⁴C₂(√3)²(√2)² + (√2)⁴]
2. = 2[9 + 6 × 3 × 2 + 4] = 2 × 49

Answer: 98

Special forms: (x − y)ⁿ and (1 + x)ⁿ

Replacing y by −y gives (x − y)ⁿ = ⁿC₀xⁿ − ⁿC₁xⁿ⁻¹y + ⁿC₂xⁿ⁻²y² − … + (−1)ⁿⁿCₙyⁿ: signs alternate, and the term with yʳ has sign (−1)ʳ. Putting x = 1: (1 + x)ⁿ = 1 + ⁿC₁x + ⁿC₂x² + … + xⁿ. Adding and subtracting (x + y)ⁿ and (x − y)ⁿ keeps only the even-position or odd-position terms, which is how surd expressions such as (√3 + √2)⁴ + (√3 − √2)⁴ become whole numbers.

Worked example

Expand (x − 2/x)⁴.

Try it first, then show the solutionSolution

1. x⁴ − 4x³(2/x) + 6x²(4/x²) − 4x(8/x³) + 16/x⁴
2. = x⁴ − 8x² + 24 − 32/x² + 16/x⁴

Answer: x⁴ − 8x² + 24 − 32/x² + 16/x⁴

The general term

The (r + 1)th term of (x + y)ⁿ is Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ yʳ, r = 0, 1, …, n. Note the shift: the 5th term has r = 4. For (x − y)ⁿ, Tᵣ₊₁ = (−1)ʳ ⁿCᵣ xⁿ⁻ʳ yʳ. When x or y carries a coefficient, raise the whole thing to the power: (3x)³ = 27x³. The ratio of consecutive terms is Tᵣ₊₁/Tᵣ = ((n − r + 1)/r)(y/x), useful for comparing neighbouring terms.

Worked example

Find the 4th term of (3x − 2)⁶.

Try it first, then show the solutionSolution

1. 4th term: r = 3; T₄ = ⁶C₃ (3x)³ (−2)³
2. = 20 × 27x³ × (−8)

Answer: −4320x³

Particular terms: coefficient of xᵏ and term independent of x

Write Tᵣ₊₁ and simplify the power of x into a single exponent in terms of r. To find the coefficient of xᵏ, set that exponent equal to k and solve for r; to find the term independent of x (the constant term), set it equal to 0. If r comes out as a non-integer or outside 0 ≤ r ≤ n, no such term exists. Then substitute r back, keeping all numerical factors and signs.

Worked example

In (x² − 2/x)⁹, find (i) the term independent of x (ii) the coefficient of x⁶.

Try it first, then show the solutionSolution

1. Tᵣ₊₁ = ⁹Cᵣ (x²)⁹⁻ʳ (−2/x)ʳ = ⁹Cᵣ (−2)ʳ x¹⁸⁻³ʳ
2. (i) 18 − 3r = 0 ⇒ r = 6: ⁹C₆ × 64 = 84 × 64 = 5376
3. (ii) 18 − 3r = 6 ⇒ r = 4: ⁹C₄ × 16 = 126 × 16 = 2016

Answer: 5376; 2016

Middle term or terms

(x + y)ⁿ has n + 1 terms. If n is even, n + 1 is odd and there is one middle term, T₍ₙ/₂₎₊₁. If n is odd, there are two middle terms, T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂. The binomial coefficient of the middle term (or terms) is the greatest coefficient in the expansion.

Worked example

Find the middle term(s) of (i) (x/3 + 3y)¹⁰ (ii) (2x − y)⁷.

Try it first, then show the solutionSolution

1. (i) n = 10: middle term T₆, r = 5: ¹⁰C₅(x/3)⁵(3y)⁵ = 252x⁵y⁵
2. (ii) n = 7: T₄ and T₅
3. T₄ = ⁷C₃(2x)⁴(−y)³ = −560x⁴y³; T₅ = ⁷C₄(2x)³(−y)⁴ = 280x³y⁴

Answer: (i) 252x⁵y⁵ (ii) −560x⁴y³ and 280x³y⁴

Applications: powers, approximations and divisibility

Large powers: write the base as a round number plus or minus a small one, e.g. 99⁴ = (100 − 1)⁴. Approximations: for small x, (1 + x)ⁿ ≈ 1 + nx + ⁿC₂x², since later terms are tiny. Divisibility: write the base as 1 + k; then (1 + k)ⁿ = 1 + nk + k²(…), so (1 + k)ⁿ − nk − 1 is divisible by k².

Worked example

Prove that 25ⁿ − 24n − 1 is divisible by 576 for every positive integer n.

Try it first, then show the solutionSolution

1. 25ⁿ = (1 + 24)ⁿ = 1 + 24n + ⁿC₂24² + ⁿC₃24³ + … + 24ⁿ
2. 25ⁿ − 24n − 1 = 24²[ⁿC₂ + ⁿC₃24 + … + 24ⁿ⁻²], and 24² = 576
3. For n = 1 the expression is 0, which is also divisible by 576

Answer: Proved

Binomial coefficients and their sums

Writing Cᵣ for ⁿCᵣ: putting x = 1 in (1 + x)ⁿ gives C₀ + C₁ + … + Cₙ = 2ⁿ; putting x = −1 gives C₀ − C₁ + C₂ − … = 0. Adding and halving: C₀ + C₂ + C₄ + … = C₁ + C₃ + C₅ + … = 2ⁿ⁻¹. The sum of the coefficients of any polynomial expansion is found by putting every variable equal to 1: for (3x − 2y)⁵ it is (3 − 2)⁵ = 1.

Worked example

Find the sum of the coefficients in (2x − 3y)⁶ and the sum C₀ + C₂ + C₄ + C₆ for n = 6.

Try it first, then show the solutionSolution

1. Put x = y = 1: (2 − 3)⁶ = 1
2. C₀ + C₂ + C₄ + C₆ = 2⁵ = 32 (1 + 15 + 15 + 1)

Answer: 1; 32

Maths in real life: Binomial Theorem

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.

Topics in this chapter

• Binomial Expansion and the General Term
• Middle Terms, Binomial Coefficients and Applications

Binomial Theorem formula sheet

Binomial theorem

$(x+y{)}^n={\sum }_{r=0}^n{}^n{C}_r{x}^{n-r}{y}^r$

n + 1 terms.

General term

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

For (x − y)ⁿ multiply by (−1)ʳ.

Middle term, n even

$T_{\frac{n}{2}+1}$

One middle term.

Middle terms, n odd

$T_{\frac{n+1}{2}},T_{\frac{n+3}{2}}$

Two middle terms.

Sum of coefficients

${}^n{C}_0+{}^n{C}_1+\cdots +{}^n{C}_n=2^n$

Even-position sum = odd-position sum = 2ⁿ⁻¹.

Ratio of consecutive terms

$\frac{T_{r+1}}{T_r}=\frac{n-r+1}{r}\cdot \frac{y}{x}$

Useful for comparing neighbouring terms.

Pascal's rule

${}^n{C}_{r-1}+{}^n{C}_r={}^{n+1}{C}_r$

Builds each row of the triangle.

Binomial Theorem: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Binomial Theorem

The question patterns to expect, and the method that works for each.

Tricks to remember Binomial Theorem

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

• Down, up, choose

  First term's power goes down, second term's power goes up, coefficient is ⁿCᵣ with r = the power going up.

  Use it when: Writing any expansion.

• Term number is one ahead

  Tᵣ₊₁: the 7th term uses r = 6.

  Use it when: General term.

• Even n, one middle

  Even power, one middle; odd power, two middles.

  Use it when: Middle terms.

• Put 1 for the sum

  Sum of coefficients: replace every variable by 1.

  Use it when: Coefficient sums.

• Bracket the whole term

  Raise (3x), (−2/x) as complete brackets so their numbers and signs are powered too.

  Use it when: Avoiding coefficient slips.

Binomial Theorem: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Binomial Theorem

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

• ✗Mistake: Using r for the rth term

  The general term is Tᵣ₊₁, not Tᵣ.

  Wrong working

  6th term of (x + 2)⁸ = ⁸C₆ x² 2⁶.

  Right working

  6th term uses r = 5: ⁸C₅ x³ 2⁵ = 1792x³.

  ✓Do this: Write 'Tᵣ₊₁, r = term number − 1' every time.

  Usually costs: Full marks for the part.

• ✗Mistake: Not powering the coefficient

  Students write (3x)³ as 3x³.

  Wrong working

  T₄ of (3x − 2)⁶ = 20 × 3x³ × (−8).

  Right working

  20 × 27x³ × (−8) = −4320x³.

  ✓Do this: Keep each term in brackets until it is powered.

  Usually costs: 1-2 marks.

• ✗Mistake: Losing the sign of a negative term

  (−y)ʳ is negative when r is odd.

  Wrong working

  T₄ of (2x − y)⁷ = +560x⁴y³.

  Right working

  (−y)³ = −y³, so T₄ = −560x⁴y³.

  ✓Do this: Write (−1)ʳ explicitly in the general term.

  Usually costs: 1 mark.

• ✗Mistake: Mishandling powers of 1/x

  Students add instead of subtract the exponent.

  Wrong working

  (x²)⁹⁻ʳ(1/x)ʳ = x¹⁸⁻ʳ.

  Right working

  x¹⁸⁻²ʳ × x⁻ʳ = x¹⁸⁻³ʳ.

  ✓Do this: Write 1/x as x⁻¹ before collecting.

  Usually costs: 1-2 marks.

• ✗Mistake: Giving one middle term for odd n

  Students compute n/2 + 1 for n = 7.

  Wrong working

  Middle term of (2x − y)⁷ is T₄.₅.

  Right working

  n odd: T₄ and T₅.

  ✓Do this: Count the terms: n + 1 = 8, so two middles.

  Usually costs: 1 mark.

Practice questions on Binomial Theorem

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

1. Q1 · 1 mark · MCQExam level

   The coefficient of x² in (1 + x)⁵ is:
   (a) 5
   (b) 10
   (c) 20
   (d) 25

   Hint

   ⁵C₂.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   The middle term of (a + b)⁸ is:
   (a) T₄
   (b) T₅
   (c) T₄ and T₅
   (d) T₉

   Hint

   n even: T₍ₙ/₂₎₊₁.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   How many terms are there in the expansion of (2x − 3y)¹¹?

   Hint

   n + 1.

   Answer

   12

4. Q4 · 1 mark · Very short answerExam level

   Write the sum of the binomial coefficients in (x + y)⁷.

   Hint

   2ⁿ.

   Answer

   128

5. Q5 · 2 marks · Short answerExam level

   Expand (a − 3)⁴.

   Hint

   Row 4 with alternating signs.

   Answer

   a⁴ − 12a³ + 54a² − 108a + 81

6. Q6 · 2 marks · Short answerHarder exam level

   Find the 5th term of (2a − b/2)⁶.

   Hint

   r = 4.

   Answer

   ⁶C₄(2a)²(−b/2)⁴ = 15 × 4 × (1/16) a²b⁴ = (15/4)a²b⁴

7. Q7 · 2 marks · Short answerHarder exam level

   Use the binomial theorem to evaluate 98⁴.

   Hint

   (100 − 2)⁴, with alternating signs.

   Answer

   100 000 000 − 8 000 000 + 240 000 − 3200 + 16 = 92 236 816

8. Q8 · 3 marks · Short answerHarder exam level

   Find the term independent of x in (x − 1/(2x))¹⁰.

   Hint

   Power of x is 10 − 2r.

   Answer

   r = 5: ¹⁰C₅(−1/2)⁵ = 252 × (−1/32) = −63/8

9. Q9 · 3 marks · Short answerHarder exam level

   Find the coefficient of x⁷ in (x² + 1/x)⁸.

   Hint

   Power of x is 16 − 3r.

   Answer

   16 − 3r = 7 ⇒ r = 3: ⁸C₃ = 56

10. Q10 · 3 marks · Short answerHarder exam level

    Find the middle term of (x² − 2/x)¹⁰.

    Hint

    T₆.

    Answer

    ¹⁰C₅(x²)⁵(−2/x)⁵ = 252 × (−32)x⁵ = −8064x⁵

11. Q11 · 3 marks · Short answerHarder exam level

    Prove that 11ⁿ − 10n − 1 is divisible by 100 for every positive integer n.

    Hint

    11 = 1 + 10.

    Answer

    11ⁿ = (1 + 10)ⁿ = 1 + 10n + 100[ⁿC₂ + ⁿC₃·10 + …], so 11ⁿ − 10n − 1 is a multiple of 100 (it is 0 when n = 1).

12. Q12 · 3 marks · Short answerHarder exam level

    The first three terms in the expansion of (1 + ax)ⁿ are 1 + 12x + 64x². Find a and n.

    Hint

    na = 12 and ⁿC₂a² = 64.

    Answer

    n(n − 1)a²/2 = 64 ⇒ (na)(n − 1)a = 128 ⇒ (n − 1)a = 32/3. Then na − a = 12 − a = 32/3 ⇒ a = 4/3, n = 9.

13. Q13 · 5 marks · Long answerHarder exam level

    (a) Prove that ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ. [1]
    (b) Prove that ⁿC₀ + ⁿC₂ + ⁿC₄ + … = ⁿC₁ + ⁿC₃ + … = 2ⁿ⁻¹. [2]
    (c) The sum of the coefficients of the odd-numbered terms (1st, 3rd, 5th, …) in (1 + x)ⁿ is 512. Find n. [2]

    Hint

    Put x = 1 and x = −1 in (1 + x)ⁿ.

    Answer

    (a) x = 1 in (1 + x)ⁿ. (b) x = −1 gives C₀ − C₁ + C₂ − … = 0, so the two alternate sums are equal; each is half of 2ⁿ. (c) Odd-numbered terms have coefficients C₀, C₂, …, summing to 2ⁿ⁻¹ = 512 = 2⁹, so n = 10.

14. Q14 · 3 marks · Short answerHOTS / competency

    Find the coefficient of x⁴ in (1 + x)⁴(1 − x)⁴.

    Hint

    Combine the brackets first.

    Answer

    (1 + x)⁴(1 − x)⁴ = (1 − x²)⁴; the x⁴ term comes from r = 2: ⁴C₂(−x²)² = 6x⁴. Coefficient 6.

15. Q15 · 3 marks · Short answerHOTS / competency

    Show that (2 + √3)⁴ + (2 − √3)⁴ = 194, and hence find the integer part of (2 + √3)⁴.

    Hint

    0 < 2 − √3 < 1.

    Answer

    Sum = 2[2⁴ + 6 × 2² × 3 + 3²] = 2[16 + 72 + 9] = 194. Since 0 < (2 − √3)⁴ < 1, (2 + √3)⁴ lies between 193 and 194, so its integer part is 193.

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. 1.General term of (x + y)ⁿ?Show answer

   Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ yʳ.

2. 2.Number of terms in (x + y)ⁿ?Show answer

   n + 1.

3. 3.Middle term when n is even?Show answer

   T₍ₙ/₂₎₊₁.

4. 4.Middle terms when n is odd?Show answer

   T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂.

5. 5.Sum of all binomial coefficients?Show answer

   2ⁿ.

6. 6.Sign of the term with yʳ in (x − y)ⁿ?Show answer

   (−1)ʳ.

7. 7.How do you find the term independent of x?Show answer

   Set the power of x in Tᵣ₊₁ equal to 0 and solve for r.

8. 8.Row 5 of Pascal's triangle?Show answer

   1, 5, 10, 10, 5, 1.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write row 7 of Pascal's triangle from memory or by building it?
• ☐Can you prove the binomial theorem by induction?
• ☐Can you expand (ax + b/x)⁴ correctly?
• ☐Can you find the 5th term of an expansion without listing the others?
• ☐Can you find the term independent of x?
• ☐Can you find the middle term(s) for n = 9 and n = 10?
• ☐Can you evaluate 99⁵ or (1.01)⁶ using the theorem?
• ☐Can you prove C₀ + C₂ + … = 2ⁿ⁻¹?

Binomial Theorem: frequently asked questions