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Lesson 2 of 3

Binomial Theorem · Lesson 2 of 3

Expanding and Applying Binomials

“Choose useful substitutions and use expansions for exact calculations, comparisons, identities, and approximations.”

Learning Objectives

• Expand binomials whose terms contain numerical factors, powers, or reciprocals. • Rewrite numbers near convenient bases to evaluate their powers exactly. • Use positivity of selected expansion terms to compare large powers. • Prove divisibility and coefficient-sum identities using suitable substitutions. • Use paired expansions to retain even or odd powers and simplify radical expressions. • Approximate using initial terms and expand three-part expressions by grouping.

Choosing the complete two terms

The theorem is an organising rule, but you still have to decide what belongs in each of its two slots. A term might be x², 3/x, or an entire bracketed expression. Write those two terms clearly before expanding, raise each complete term to its assigned power, and only then simplify the powers and numerical factors.

For a reciprocal expression, the original domain restriction remains essential. An expansion of x² + 3/x is meaningful only for x ≠ 0. Simplifying the result does not grant permission to use a value that made the original expression undefined. Another useful check is to keep coefficient, sign, and exponent calculations separate until the end.

Example — A reciprocal inside a binomial

Problem
Expand (x² + 3/x)⁴, where x ≠ 0.

  1. 1.Choose A = x² and B = 3/x. Use the fourth-power row 1, 4, 6, 4, 1.
  2. 2.The terms are A⁴ + 4A³B + 6A²B² + 4AB³ + B⁴.
  3. 3.A⁴ = x⁸. The next term is 4x⁶(3/x) = 12x⁵.
  4. 4.The third term is 6x⁴(9/x²) = 54x². The fourth is 4x²(27/x³) = 108/x. The final term is 81/x⁴.
  5. 5.Therefore the expansion is x⁸ + 12x⁵ + 54x² + 108/x + 81/x⁴, still with x ≠ 0.
  6. 6.The resulting exponents of x are 8, 5, 2, −1, −4. They decrease by 3 because the first slot contributes power 2 and the second slot contributes power −1.
Example — Subtraction with reciprocal terms

Problem
Expand (2/x − x/2)⁵ for x ≠ 0.

  1. 1.Use A = 2/x and B = −x/2 with the coefficient row 1, 5, 10, 10, 5, 1.
  2. 2.The first two terms are A⁵ = 32/x⁵ and 5A⁴B = 5(16/x⁴)(−x/2) = −40/x³.
  3. 3.The next two are 10A³B² = 10(8/x³)(x²/4) = 20/x and 10A²B³ = 10(4/x²)(−x³/8) = −5x.
  4. 4.The last two are 5AB⁴ = 5(2/x)(x⁴/16) = 5x³/8 and B⁵ = −x⁵/32.
  5. 5.The result is 32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32. Check the alternating signs before collecting the answer.
A denominator is also raised to its power

For instance, (3/x)² is 9/x², not 9/x. A negative term has a sign factor (−1)ʳ at position r + 1. Dropping brackets too soon can hide either of these facts. Preserve x ≠ 0 whenever a reciprocal occurs in the original expression.

Exact numerical evaluation near a convenient base

A number close to 100 or 1000 can be written as a convenient base plus a small adjustment. Powers of the base are easy to calculate, and the theorem combines them with powers of the adjustment. Since the exponent is a positive integer, retaining every term gives an exact value; no approximation is involved.

Example — An exact fifth power

Problem
Use a binomial expansion to calculate 98⁵.

  1. 1.Write 98 = 100 − 2 and use the fifth-power coefficients 1, 5, 10, 10, 5, 1.
  2. 2.Expand as 100⁵ − 5(100⁴)(2) + 10(100³)(2²) − 10(100²)(2³) + 5(100)(2⁴) − 2⁵.
  3. 3.The six numerical terms are 10,000,000,000; −1,000,000,000; 40,000,000; −800,000; 8,000; and −32.
  4. 4.Add in stages: 10,000,000,000 − 1,000,000,000 = 9,000,000,000; then add 40,000,000 and subtract 800,000 to get 9,039,200,000.
  5. 5.Adding 8,000 and subtracting 32 gives 98⁵ = 9,039,207,968. The value is below 100⁵, as expected.
Example — A small positive adjustment

Problem
Calculate 102⁵ using a binomial expansion.

  1. 1.Write 102 = 100 + 2. All six expansion contributions are positive.
  2. 2.Use 100⁵ + 5(100⁴)(2) + 10(100³)(2²) + 10(100²)(2³) + 5(100)(2⁴) + 2⁵.
  3. 3.These are 10,000,000,000 + 1,000,000,000 + 40,000,000 + 800,000 + 8,000 + 32.
  4. 4.Their sum is 11,040,808,032. The adjustment terms explain why the answer exceeds 100⁵.
  5. 5.Compare this with 98⁵: the even-adjustment contributions have the same sign, while the odd-adjustment contributions switch sign.

The same method evaluates 96³ by taking 96 = 100 − 4, 101⁴ by taking 101 = 100 + 1, and 99⁵ by taking 99 = 100 − 1. Choose the representation that makes the powers easiest, rather than always choosing the smallest possible base. A check on the overall size can catch an arithmetic error even before every digit is recomputed.

Comparing large powers without calculating them

Sometimes an exact decimal expansion would be enormous, and the question only asks which quantity is larger. For a positive adjustment t, every term in (1 + t)ⁿ is positive. The first two terms are therefore a lower bound, and the remaining terms make the full value larger when n ≥ 2. This works because the sign of every omitted term is known.

A lower bound from positive termsLaTeX
n is an integer. For n = 1 the equality (1 + t) = 1 + t holds, so the strict bound shown here needs n ≥ 2.
Example — A millionth power

Problem
Which is larger: (1.01)¹⁰⁰⁰⁰⁰⁰ or 10,000?

  1. 1.Write 1.01 = 1 + 0.01. The first two terms of the expansion are 1 and 1,000,000 × 0.01 = 10,000.
  2. 2.Every later term C(1,000,000, r)(0.01)ʳ is positive.
  3. 3.Thus the full value is greater than 1 + 10,000 = 10,001, which is already greater than 10,000.
  4. 4.Therefore (1.01)¹⁰⁰⁰⁰⁰⁰ is larger. No exact evaluation of the millionth power is necessary.
Example — A second comparison

Problem
Compare (1.1)¹⁰⁰⁰⁰ with 1000.

  1. 1.Use 1.1 = 1 + 0.1 and the first two terms 1 + 10,000(0.1) = 1001.
  2. 2.All remaining terms are positive because 0.1 is positive.
  3. 3.Hence (1.1)¹⁰⁰⁰⁰ > 1001 > 1000. This is an exact inequality, even though we did not calculate the full value.
Omitting terms needs a reason

A shortened expansion is not automatically a lower bound. The positivity argument above works for a positive adjustment. In (1 − t)ⁿ the signs alternate, so the omitted terms cannot all be assumed positive. Use the actual signs or calculate the remainder before claiming the direction of an estimate.

Remainders and divisibility

A divisibility question asks whether an expression is an integer multiple of a particular number. A remainder question asks whether it can be written as that multiple plus a specified small value. Binomial expansions are useful because the initial terms can be separated and every later term may contain a common factor.

If an integer N can be written as N = dq + r with integer q and 0 ≤ r < d, then division by positive d leaves remainder r. To prove a remainder of 1 on division by 25, we seek a form 25q + 1. The important point is that q is an integer; q may be zero in a small boundary case.

Example — A remainder of one

Problem
Prove that 6ⁿ − 5n leaves remainder 1 when divided by 25 for every positive integer n.

  1. 1.Write 6ⁿ = (1 + 5)ⁿ. Its expansion starts 1 + 5n, followed by terms C(n, r)5ʳ for r ≥ 2.
  2. 2.Subtract 5n to obtain 6ⁿ − 5n = 1 + Σ from r = 2 to n of C(n, r)5ʳ.
  3. 3.For every r ≥ 2, factor 25 from 5ʳ. This gives 6ⁿ − 5n = 1 + 25K, where K = Σ from r = 2 to n of C(n, r)5ʳ⁻² is an integer.
  4. 4.For n = 1 there are no later terms: K = 0 and 6 − 5 = 1. For n ≥ 2 the same factorisation applies to the displayed later terms.
  5. 5.Therefore the remainder upon division by 25 is always 1. This concerns 6ⁿ − 5n, not 6ⁿ − 5ⁿ.
Example — A divisibility identity

Problem
Show that 9ⁿ⁺¹ − 8n − 9 is divisible by 64 whenever n is a positive integer.

  1. 1.Rewrite 9ⁿ⁺¹ as (1 + 8)ⁿ⁺¹.
  2. 2.The first two expansion terms are 1 + 8(n + 1) = 8n + 9.
  3. 3.Subtracting 8n + 9 cancels those terms, leaving Σ from r = 2 to n + 1 of C(n + 1, r)8ʳ.
  4. 4.Each remaining term contains 8² = 64. Factoring it leaves an integer sum, so the expression is divisible by 64.
  5. 5.At n = 1, the expression is 81 − 8 − 9 = 64, consistent with the argument.
Example — A factor of a difference of powers

Problem
For distinct integers a and b, prove that a − b divides aⁿ − bⁿ for positive integer n.

  1. 1.Write a = b + (a − b). Choose the first binomial term to be b and the second to be a − b, so aⁿ = [b + (a − b)]ⁿ.
  2. 2.The expansion is Σ from r = 0 to n of C(n, r)bⁿ⁻ʳ(a − b)ʳ. Its r = 0 term is bⁿ; every term with r ≥ 1 contains at least one factor a − b.
  3. 3.Subtract bⁿ. The remaining sum is Σ from r = 1 to n of C(n, r)bⁿ⁻ʳ(a − b)ʳ.
  4. 4.Factor a − b from the entire sum. The quotient is Σ from r = 1 to n of C(n, r)bⁿ⁻ʳ(a − b)ʳ⁻¹, which is an integer.
  5. 5.Thus aⁿ − bⁿ is an integer multiple of a − b. Distinctness guarantees that the divisor is nonzero; for n = 1 the quotient is 1.

Substitution identities and paired expansions

The theorem is an identity, so substituting suitable values gives new identities. Choosing a second term of 3 produces a weighted coefficient sum. Comparing expansions with opposite second terms produces cancellations: adding retains the even powers of that term, while subtracting retains the odd powers. This is especially useful with square roots.

Example — A weighted coefficient sum

Problem
Prove that Σ from r = 0 to n of 3ʳC(n, r) equals 4ⁿ.

  1. 1.Use (1 + x)ⁿ = Σ from r = 0 to n of C(n, r)xʳ.
  2. 2.Set x = 3. The left side becomes (1 + 3)ⁿ = 4ⁿ.
  3. 3.The right side becomes exactly the required sum C(n, 0) + 3C(n, 1) + 3²C(n, 2) + … + 3ⁿC(n, n).
  4. 4.Therefore the sum is 4ⁿ. This is the same substitution principle that produced the ordinary coefficient sum when x = 1.

Consider the matching r positions in (A + B)ⁿ and (A − B)ⁿ. Their contributions have equal magnitude but differ by a factor (−1)ʳ. For even r the terms agree, so addition doubles them and subtraction removes them. For odd r the signs oppose, so addition removes them and subtraction doubles them. Keeping the same term order makes this cancellation visible.

What addition and subtraction retainLaTeX
The corresponding difference keeps twice the terms with odd r. The parity concerns r, the exponent of B, not the position number r + 1.
Example — A fourth-power difference

Problem
Find (a + b)⁴ − (a − b)⁴, then evaluate it at a = √3 and b = √2.

  1. 1.The plus expansion is a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
  2. 2.The minus expansion is a⁴ − 4a³b + 6a²b² − 4ab³ + b⁴.
  3. 3.Subtract the entire second expansion: the even-b terms cancel and the odd-b terms double, giving 8a³b + 8ab³ = 8ab(a² + b²).
  4. 4.At a = √3 and b = √2, ab = √6 and a² + b² = 5.
  5. 5.The value is 8√6 × 5 = 40√6. Subtracting a bracket requires changing every sign in that bracket.
Example — A sixth-power sum

Problem
Find (x + 1)⁶ + (x − 1)⁶, then evaluate at x = √2.

  1. 1.The sixth-power coefficients are 1, 6, 15, 20, 15, 6, 1.
  2. 2.Addition retains the terms with even powers of the second term 1: 2(x⁶ + 15x⁴ + 15x² + 1).
  3. 3.Thus the polynomial is 2x⁶ + 30x⁴ + 30x² + 2.
  4. 4.At x = √2, x² = 2, x⁴ = 4, and x⁶ = 8. The value is 2(8 + 60 + 30 + 1) = 198.
  5. 5.We did not need to evaluate the individual sixth powers of the two irrational expressions.
Example — A sixth-power difference of conjugate expressions

Problem
Evaluate (√3 + √2)⁶ − (√3 − √2)⁶.

  1. 1.Set A = √3 and B = √2. Subtraction keeps twice the positions with r = 1, 3, and 5.
  2. 2.The expression is 12A⁵B + 40A³B³ + 12AB⁵.
  3. 3.Factor AB = √6: the bracket is 12A⁴ + 40A²B² + 12B⁴.
  4. 4.Since A² = 3 and B² = 2, the bracket is 12 × 9 + 40 × 6 + 12 × 4 = 396.
  5. 5.The exact result is 396√6. Keeping only the odd-B terms made the radical arithmetic much smaller.
Example — Keeping a squared first term intact

Problem
Simplify (a² + √(a² − 1))⁴ + (a² − √(a² − 1))⁴ over the real numbers.

  1. 1.For real square roots, require a² − 1 ≥ 0. Let A = a² and B = √(a² − 1).
  2. 2.The fourth-power sum is 2(A⁴ + 6A²B² + B⁴), because the odd-B terms cancel.
  3. 3.Substitute carefully: A⁴ = a⁸, A²B² = a⁴(a² − 1), and B⁴ = (a² − 1)².
  4. 4.The result is 2[a⁸ + 6a⁴(a² − 1) + (a² − 1)²].
  5. 5.Expanding the brackets gives 2a⁸ + 12a⁶ − 10a⁴ − 4a² + 2, valid for a² ≥ 1.
  6. 6.The first slot is a², not a. Keeping that distinction prevents an incorrect simplification.

Approximating with the initial terms

For a small adjustment, later powers of the adjustment are often much smaller than the early ones. Using only the first few terms can then give a useful approximation. The theorem is still an exact finite identity; the approximation arises because we deliberately omit some of its terms. We should name the omitted contribution and check whether its size suits the purpose.

Example — A first-three-term approximation

Problem
Approximate (0.99)⁵ using the first three terms of its expansion.

  1. 1.Write 0.99 = 1 − 0.01. The exact expansion is 1 − 5(0.01) + 10(0.01)² − 10(0.01)³ + 5(0.01)⁴ − (0.01)⁵.
  2. 2.Retain the first three terms: 1 − 0.05 + 0.001 = 0.951. Therefore (0.99)⁵ ≈ 0.951.
  3. 3.The omitted part is −0.000010 + 0.00000005 − 0.0000000001 = −0.0000099501.
  4. 4.The exact value is 0.9509900499. The approximation is larger by 0.0000099501.
  5. 5.This shows why ≈ is needed. The three-term value happens to agree with the exact value rounded to three decimal places, but truncation and rounding are different operations.
An approximation needs its own symbol

Do not write (0.99)⁵ = 0.951. The exact expansion contains six terms, and the retained three have a nonzero remainder. An estimate using positive omitted terms can be a lower bound; an alternating remainder needs its actual signs checked.

A three-part expression can be grouped into two parts

The theorem applies to two slots, but each slot can contain a whole expression. To expand a three-part expression, group two of its parts into one slot. Expand the outer binomial, then expand the remaining bracketed powers. This uses the binomial theorem repeatedly and does not require a new theorem for three terms.

Example — A grouped reciprocal expression

Problem
Expand (1 + x/2 − 2/x)⁴, with x ≠ 0.

  1. 1.Put u = x/2 − 2/x. The outer expansion is (1 + u)⁴ = 1 + 4u + 6u² + 4u³ + u⁴.
  2. 2.Let A = x/2 and B = −2/x. Then u = A + B and AB = −1. The square is u² = A² + 2AB + B² = x²/4 − 2 + 4/x².
  3. 3.For the cube, use u³ = A³ + 3A²B + 3AB² + B³. The four contributions are x³/8, −3x/2, 6/x, and −8/x³, respectively.
  4. 4.For the fourth power, use u⁴ = A⁴ + 4A³B + 6A²B² + 4AB³ + B⁴. The five contributions are x⁴/16, −x², 6, −16/x², and 16/x⁴. In particular, A²B² = (AB)² = 1, which explains the constant 6.
  5. 5.Now multiply u, u², and u³ by their outer coefficients: 4u = 2x − 8/x; 6u² = 3x²/2 − 12 + 24/x²; 4u³ = x³/2 − 6x + 24/x − 32/x³.
  6. 6.Combine these with 1 and u⁴. The result is x⁴/16 + x³/2 + x²/2 − 4x − 5 + 16/x + 8/x² − 32/x³ + 16/x⁴.
  7. 7.Every occurrence of u used the same definition. The restriction x ≠ 0 remains part of the answer.
Example — A grouped polynomial cube

Problem
Expand (3x² − 2ax + 3a²)³ using the binomial theorem.

  1. 1.Choose U = 3x² + 3a² and V = −2ax. The outer cube is U³ + 3U²V + 3UV² + V³.
  2. 2.U³ = 27(x² + a²)³ = 27x⁶ + 81a²x⁴ + 81a⁴x² + 27a⁶.
  3. 3.3U²V = −54ax(x² + a²)² = −54ax⁵ − 108a³x³ − 54a⁵x.
  4. 4.3UV² = 36a²x²(x² + a²) = 36a²x⁴ + 36a⁴x², and V³ = −8a³x³.
  5. 5.Collect like powers of x. The expansion is 27x⁶ − 54ax⁵ + 117a²x⁴ − 116a³x³ + 117a⁴x² − 54a⁵x + 27a⁶.
  6. 6.As a check, a = 0 leaves 27x⁶, agreeing with (3x²)³. The paired coefficients also agree with the symmetry between x and a in the original expression.

Check your understanding

The same expansion can answer several different questions, but the most efficient use depends on the goal. Decide whether you need every term, a useful bound, a common factor, or cancellation between two matching expansions before choosing an answer.

Quiz

Quick check

What is the term with r = 2 in (x² + 3/x)⁴?

Quick check

Which rewriting is convenient for calculating 99⁵ exactly?

Quick check

Why do the first two terms prove (1.1)¹⁰⁰⁰⁰ > 1000?

Quick check

For n ≥ 2, after cancelling 1 + 5n in (1 + 5)ⁿ, what common factor occurs in all remaining terms?

Quick check

What does (A + B)ⁿ + (A − B)ⁿ retain?

Quick check

Which substitution proves Σ from r = 0 to n of 3ʳC(n, r) = 4ⁿ?

Quick check

Which statement about the three-term approximation of (0.99)⁵ is correct?

Quick check

How can the binomial theorem be used on (P + Q + R)⁴?

Practice Problems

Practice Problems
  1. Expand (1 − 2x)⁵ and (2x − 3)⁶, showing the origin of each sign and numerical factor.
  2. Expand (x/3 + 1/x)⁵ and (x + 1/x)⁶, keeping their domain restrictions.
  3. Use convenient bases to calculate 96³, 101⁴, and 99⁵ exactly.
  4. Explain why the positive-term comparison works for (1.1)¹⁰⁰⁰⁰, and why the same explanation cannot simply be reused for (0.9)¹⁰⁰⁰⁰.
  5. Derive (a + b)⁴ − (a − b)⁴ before evaluating it at a = √3 and b = √2.
  6. Derive (x + 1)⁶ + (x − 1)⁶ and evaluate it at x = √2.
  7. Rebuild the proof that 9ⁿ⁺¹ − 8n − 9 is divisible by 64. Identify the cancelled terms.
  8. Prove the weighted coefficient identity Σ from r = 0 to n of 3ʳC(n, r) = 4ⁿ by substitution.
  9. Explain why a − b is a factor of aⁿ − bⁿ for distinct integers a and b.
  10. Use cancellation to evaluate (√3 + √2)⁶ − (√3 − √2)⁶ without finding the two powers separately.
  11. Simplify (a² + √(a² − 1))⁴ + (a² − √(a² − 1))⁴, and state its real domain.
  12. Find the first-three-term approximation to (0.99)⁵, then calculate the omitted contribution.
  13. Re-expand (1 + x/2 − 2/x)⁴ by grouping the two reciprocal terms first.
  14. Expand (3x² − 2ax + 3a²)³ and check the result at a = 0 and x = 0.

Key Takeaways

Key Takeaways

• Choose and bracket the complete binomial terms before applying their powers. • A convenient base turns an exact numerical power into a manageable finite sum. • Known positive omitted terms justify a lower bound; omitted terms of mixed signs need further checking. • Divisibility proofs cancel initial terms and expose a common integer factor in the rest. • Adding paired plus/minus expansions keeps even second-term powers; subtracting keeps odd powers. • Initial-term approximations need ≈ and an awareness of the remainder. • Grouping allows the same theorem to expand expressions with three parts.

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Discovering the Binomial Theorem

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Chapter Summary and Practice