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Lesson 3 of 3

Binomial Theorem · Lesson 3 of 3

Chapter Summary and Practice

“Connect the coefficient pattern, proof, and applications, then choose and check methods in mixed problems.”

Learning Objectives

• Connect Pascal’s triangle, combinations, the general expansion, and its induction proof. • Track coefficients, powers, position numbers, and signs accurately. • Select substitutions for numerical calculations and coefficient identities. • Distinguish full evaluation, a justified bound, and an approximation. • Use cancellation, common factors, and grouping in mixed applications. • Check expansions using domains, substitutions, and omitted-term reasoning.

One structure behind the whole chapter

The chapter begins with a visible coefficient pattern and develops it into a formula that works for every positive integer power. That formula then becomes a tool for numerical evaluation, identities, comparison, divisibility, and approximation. The methods are connected by one question: which terms of an expansion do we need, and why? A full calculation needs them all; a bound, a cancellation, or a factor proof may need only their structure.

Start each problem by identifying the exponent and the complete two terms. Decide whether the goal is an expansion, a value, an inequality, or an identity. Write the needed terms with their coefficients and powers, simplify carefully, and check the result. If the expression has three parts, group it into two slots first. If it contains reciprocals or real square roots, state the domain before doing algebra.

Chapter ideaCore relationshipUseful check
Powers and patternsn + 1 standard positions; first-slot exponent falls as second-slot exponent risesThe two placeholder exponents sum to n
Pascal’s triangleEndpoints are 1; interior entries are adjacent sumsLabel rows by n, beginning with n = 0
CombinationsC(n, r) = n!/[r!(n − r)!]0 ≤ r ≤ n and r is an integer
Coefficient meaningChoose the r factors that contribute the second termDistinguish a binomial coefficient from a coefficient after substitution
Summation notationSum C(n, r)Aⁿ⁻ʳBʳ from r = 0 to nPosition r + 1 uses index r
InductionBase case plus validity at k implying validity at k + 1Collect both contributions to each interior power
A negative second term(−B)ʳ introduces (−1)ʳOdd r changes sign; even r does not
Special forms(1 ± x)ⁿ exposes ascending powers of xUse substitution to obtain coefficient sums
Exact numerical powersChoose a convenient base and retain every termCompare the answer with the base power
Large-power comparisonPositive later terms give a justified lower boundDo not assume positivity for a negative adjustment
Remainder and factor proofsCancel initial terms and factor every remaining termThe quotient must be an integer
Paired expansionsAddition retains even r; subtraction retains odd rApply the operation to the entire second expansion
ApproximationRetain selected early terms and examine the remainderUse ≈ rather than =
Three-part expressionsGroup into two slots and expand repeatedlyKeep the grouping definition consistent
Domains and checkingOriginal restrictions remain after expansionTest simple permitted substitutions

Recovering the theorem from its meaning

A coefficient row is not a collection of unrelated numbers. Every coefficient counts the ways a certain product can arise from multiplying the factors a + b. Pascal’s rule records how those choices divide into two cases, and the induction proof records how the same contributions combine during multiplication. Remembering this connection makes the formula easier to rebuild if a coefficient or exponent is forgotten.

The central expansionLaTeX
n is a positive integer; r runs through 0, 1, …, n. A and B denote the complete two terms. Use 0! = 1 at the endpoints.

In the proof, P(1) is checked directly. Assuming the formula at k, multiplication by A produces one set of contributions and multiplication by B produces another. At each interior power, the old coefficients C(k, r) and C(k, r − 1) add to C(k + 1, r). The two endpoints each have coefficient 1. This proves the next case, allowing the conclusion to pass through every positive integer.

Example — Rebuilding a coefficient row

Problem
Find the coefficient row for power 6 and explain why the central coefficient is 20.

  1. 1.The row for power 5 is 1, 5, 10, 10, 5, 1. Put 1 at the two ends of the next row.
  2. 2.Adjacent sums give 6, 15, 20, 15, and 6, so the sixth-power row is 1, 6, 15, 20, 15, 6, 1.
  3. 3.The central coefficient is 10 + 10 = 20, or directly C(6, 3) = 6!/(3!3!) = 20.
  4. 4.It counts the ways to choose the three factors contributing B when the product is A³B³. The agreement of counting and adjacent addition links the triangle to the theorem.

Checking powers, signs, and positions

The placeholder pattern is simple, but substituted expressions can make the final answer look very different. Keep the first and second slots distinct until their powers have been evaluated. A position label, a binomial coefficient, and a final coefficient are three different pieces of information; treating them as interchangeable is a common source of errors.

Example — One position in a more complicated binomial

Problem
Find the third expansion position of (2x² − 1/x)⁵ for x ≠ 0.

  1. 1.The third position uses r = 2, because the first uses r = 0.
  2. 2.The required term is C(5, 2)(2x²)³(−1/x)².
  3. 3.C(5, 2) = 10, (2x²)³ = 8x⁶, and (−1/x)² = 1/x².
  4. 4.Multiplication gives 80x⁴. The binomial coefficient is 10, while the final coefficient is 80.
  5. 5.The sign is positive because the second slot is squared. The placeholder exponents are 3 and 2; the final x exponent is 6 − 2 = 4.
Check three independent details

A correct coefficient does not guarantee a correct term. Check the position index r, the powers of the complete slots, and the sign from any negative second term. Also distinguish the formal n + 1 expansion positions from the number of nonzero terms after substitution and simplification.

A quick domain check can prevent an invalid conclusion. For (x + 1/x)⁶ we must keep x ≠ 0, even if some of the resulting powers happen to be positive. For a real expression involving √(a² − 1), use a² ≥ 1. A test value must belong to the original domain; plugging in a forbidden value cannot validate the expansion.

Choosing substitutions and cancellations

Setting the first slot to 1 exposes a polynomial whose coefficients are combinations. Setting the second slot to a convenient value evaluates a whole weighted coefficient sum at once. Pairing two expansions with opposite second terms instead makes unwanted powers cancel. Both methods are ways to use the identity strategically rather than multiplying out everything without a goal.

Ordinary and alternating coefficient sumsLaTeX
These follow from (1 + x)ⁿ with x = 1 and x = −1 respectively. More generally, substituting x = t gives the weighted sum Σ C(n, r)tʳ = (1 + t)ⁿ.
Example — A weighted sum and a parity check

Problem
Evaluate Σ from r = 0 to 5 of 2ʳC(5, r), then find the total of the even-indexed coefficients in the row for n = 5.

  1. 1.Set x = 2 in (1 + x)⁵. The weighted sum equals 3⁵ = 243.
  2. 2.For the parity question, let E be the sum of C(5, r) with even r and O the sum with odd r.
  3. 3.The ordinary sum gives E + O = 2⁵ = 32. The alternating sum gives E − O = 0.
  4. 4.Adding these equations gives 2E = 32, so E = 16; similarly O = 16.
  5. 5.The first problem weights each coefficient by 2ʳ. The second uses the ordinary coefficients and separates them by parity; they are different sums.
Example — Simplifying before substituting radicals

Problem
Simplify (A + B)⁴ + (A − B)⁴ and evaluate it at A = √5 and B = √2.

  1. 1.The fourth-power coefficients are 1, 4, 6, 4, 1. Addition removes the odd powers of B.
  2. 2.The sum is 2(A⁴ + 6A²B² + B⁴).
  3. 3.Use A² = 5 and B² = 2: the bracket becomes 25 + 6 × 5 × 2 + 4 = 89.
  4. 4.The value is 178. Finding the two irrational fourth powers separately would create terms that eventually cancel anyway.

In a paired difference, remember that subtracting the second expansion changes every one of its signs. A negative odd-power term in that expansion becomes positive when subtracted, so the corresponding odd term doubles. Align matching powers to make this visible. In a paired sum, even second-term powers survive regardless of whether n itself is odd or even.

Exact values, bounds, and approximations

These three goals use the same finite expansion in different ways. An exact numerical answer adds every term. A bound uses the signs of omitted terms to guarantee one quantity is larger or smaller. An approximation deliberately leaves a remainder, whose size tells us whether the estimate is useful. State which kind of conclusion you have reached.

Example — An exact value near 100

Problem
Calculate 101⁴ exactly.

  1. 1.Write 101 = 100 + 1 and use the row 1, 4, 6, 4, 1.
  2. 2.The expansion is 100⁴ + 4(100³) + 6(100²) + 4(100) + 1.
  3. 3.The contributions are 100,000,000 + 4,000,000 + 60,000 + 400 + 1.
  4. 4.Their sum is 104,060,401. Every term is included, so this is an equality.
Example — A comparison from a short part of the expansion

Problem
Show that (1.02)¹⁰⁰⁰ exceeds 20.

  1. 1.Rewrite the expression as (1 + 0.02)¹⁰⁰⁰. The first two terms are 1 + 1000(0.02) = 21.
  2. 2.All later terms are positive, so the full value is greater than 21.
  3. 3.Since 21 > 20, the required comparison follows without finding an exact value. The reason for omitting the remaining terms is their known positivity.
Example — A short approximation with a visible remainder

Problem
Approximate (1.01)⁴ using its first three terms and determine the direction of the error.

  1. 1.Use (1 + 0.01)⁴ = 1 + 4(0.01) + 6(0.01)² + 4(0.01)³ + (0.01)⁴.
  2. 2.The retained terms give 1 + 0.04 + 0.0006 = 1.0406.
  3. 3.The omitted remainder is 0.000004 + 0.00000001 = 0.00000401, which is positive.
  4. 4.Thus the approximation is slightly below the exact value 1.04060401. Write (1.01)⁴ ≈ 1.0406 rather than claiming equality.
The signs of omitted terms control the claim

For a positive adjustment, omitted terms are positive and an initial partial sum is below the full value. For a negative adjustment, signs alternate. In the source example (0.99)⁵, the first-three-term value 0.951 is above the exact value because its total omitted remainder is negative. Check the actual remainder instead of guessing from the number of retained terms.

Using structure in proofs and grouped expansions

A factor proof needs an expression that displays the proposed factor in every remaining term. A remainder proof separates the small remainder from an integer multiple of the divisor. A grouped expansion treats a whole subexpression as one slot and expands it later. In all three situations, choosing a useful rewriting is the main reasoning step.

Example — A remainder proof with a new base

Problem
Show that 11ⁿ − 10n leaves remainder 1 on division by 100 for positive integer n.

  1. 1.Use 11ⁿ = (1 + 10)ⁿ. Its first two terms are 1 + 10n.
  2. 2.After subtracting 10n, the expression is 1 plus terms C(n, r)10ʳ with r ≥ 2.
  3. 3.Every later term is a multiple of 100, so 11ⁿ − 10n = 1 + 100K for an integer K.
  4. 4.For n = 1 the later sum is empty and K = 0. The remainder is 1 in every case.
Example — Grouping and checking a three-part square

Problem
Expand (x² + x + 1)² using a binomial grouping.

  1. 1.Group as [(x² + 1) + x]². Use U = x² + 1 and V = x.
  2. 2.The outer square is U² + 2UV + V².
  3. 3.Expand U² = x⁴ + 2x² + 1 and 2UV = 2x³ + 2x. Add V² = x².
  4. 4.The result is x⁴ + 2x³ + 3x² + 2x + 1.
  5. 5.At x = 1 the original value is 3² = 9, while the coefficient sum of the expansion is also 9. This check can catch errors but does not replace the derivation.

The chapter’s harder grouped examples use the same sequence at higher powers. For 1 + x/2 − 2/x, group the reciprocal difference as u and expand (1 + u)⁴. For 3x² − 2ax + 3a², grouping the symmetric terms 3x² + 3a² makes the subsequent cube organised. Write each grouping definition once and use it consistently throughout the calculation.

Check your understanding

The questions below revisit the chapter from several directions. Decide which method is justified before calculating: a coefficient count, a substitution, a cancellation, a sign argument, or a common factor. Explain the reasoning for any answer you initially find uncertain.

Quiz

Quick check

What is the coefficient of x² in the expansion of (1 + 3x)⁴?

Quick check

Which expression is the second expansion position of (A + B)⁸?

Quick check

What is Σ from r = 0 to 4 of C(4, r)(−1)ʳ?

Quick check

Which powers of B survive in (A + B)⁵ − (A − B)⁵?

Quick check

For real expressions involving √(a² − 1), which restriction is required?

Quick check

A first-three-term sum for (1 + 0.02)⁵ omits positive terms. What follows?

Quick check

Why does (1 + 8)ⁿ⁺¹ help prove divisibility of 9ⁿ⁺¹ − 8n − 9 by 64?

Quick check

Which statement correctly describes the induction step?

Quick check

Which grouping uses the binomial theorem correctly for P + Q + R?

Practice Problems

Practice Problems
  1. Build the coefficient row for n = 7 in two ways: adjacent addition and combinations. Explain why they agree.
  2. Write the binomial theorem in expanded and summation forms, defining each symbol and the allowed indices.
  3. Explain the base case, induction assumption, and coefficient-collection step of the proof without copying the final formula alone.
  4. Expand (3x − 2y)⁴ and check the coefficient and exponent of the term containing x²y².
  5. Expand (x² + 2/x)⁴ for x ≠ 0. Compare its final x exponents with the placeholder exponents.
  6. Find the fourth expansion position of (2x + 1)⁶. Distinguish its index, its binomial coefficient, and its final coefficient.
  7. Use a convenient base to evaluate 96³ and 99⁵ exactly.
  8. Use a binomial expansion to prove (1.05)²⁰⁰ > 10 without calculating its complete value.
  9. Prove Σ from r = 0 to n of 3ʳC(n, r) = 4ⁿ, then explain how the same method produces the ordinary and alternating sums.
  10. Derive the fourth-power sum and difference of A + B and A − B. Use them to evaluate paired expressions with A = √3 and B = √2.
  11. Evaluate (√3 + √2)⁶ − (√3 − √2)⁶ by retaining only the appropriate terms.
  12. Approximate (0.99)⁵ using three terms, find its exact omitted contribution, and state the direction of the error.
  13. Prove that 6ⁿ − 5n has remainder 1 on division by 25, including the case n = 1.
  14. Prove that 9ⁿ⁺¹ − 8n − 9 is divisible by 64 and that a − b divides aⁿ − bⁿ for distinct integers a and b.
  15. Simplify (a² + √(a² − 1))⁴ + (a² − √(a² − 1))⁴, stating its real domain.
  16. Expand (1 + x/2 − 2/x)⁴ by grouping and state the restriction on x.
  17. Expand (3x² − 2ax + 3a²)³, then check your result at a = 0, x = 0, and a = x = 1.
  18. A student writes (A − B)⁴ = A⁴ − 4A³B − 6A²B² − 4AB³ − B⁴. Identify every sign error and explain it using powers of −B.

Key Takeaways

Key Takeaways

• The triangle, combination counts, and induction proof describe the same coefficient structure. • The binomial theorem applies to complete two-term slots; bracket and raise those terms before simplifying. • Term position, binomial coefficient, and coefficient after substitution must be distinguished. • Substitution creates coefficient-sum identities; pairing plus/minus expansions creates parity cancellation. • Exact evaluation, a justified bound, and an approximation require different treatment of omitted terms. • Remainders and divisibility follow by cancelling initial terms and exposing common integer factors. • Grouping expands three-part expressions with the same theorem, and original domains remain in force.

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Expanding and Applying Binomials

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