MathematicsCore22 min read

Mathematical Induction and the Binomial Theorem

Proving infinitely many statements at once, and expanding a bracket without multiplying it out

This topic appears in:

1st Year Mathematics · ch.8A Level Mathematics (9709) · ch.3
01

Why induction is needed at all

Check that 1 + 3 + 5 + … + (2n − 1) = n² for n = 1, 2, 3 and 4 and it works every time. That is evidence, not proof — there are infinitely many cases left and no amount of checking will finish them.

Mathematical induction is the tool that closes the gap. It proves a statement for every natural number by proving just two things: that it holds at the start, and that whenever it holds for one number it must hold for the next.

The standard picture is a row of dominoes. Knock the first one over, and show that any falling domino knocks over the one after it, and the whole infinite row goes down. Miss either half and nothing follows.

Step 1 (basis): show P(1) is trueStep 2 (hypothesis): assume P(k) is true for some kStep 3 (inductive step): prove P(k + 1) followsConclusion: by induction, P(n) holds for all n ≥ 1all four lines are marked — the conclusion is not optional decoration

The two halves each carry marks, and each fails on its own

Without the basis, the induction step proves only "if it ever holds, it keeps holding" — and it might never hold. Without the inductive step, checking P(1) proves one case. Examiners routinely give a statement that is false to catch students who write the ritual without checking it, so do the basis honestly.

02

Writing an induction proof that gets full marks

The proof has a fixed shape and the marks are attached to its parts. Write the statement P(n) explicitly, do the basis, state the assumption in full, and in the inductive step aim deliberately at the expression you want to reach.

The single most useful habit: at the start of the inductive step, write down what P(k+1) claims, by substituting k+1 into the formula. That gives you a target, and the algebra then has somewhere to go.

Worked example

Prove by induction that 1 + 2 + 3 + … + n = n(n + 1)/2 for all n ≥ 1.

  1. Basis. For n = 1 the left side is 1 and the right side is 1(2)/2 = 1. They agree, so P(1) is true.Always evaluate both sides separately. Writing "obviously true" earns nothing.
  2. Hypothesis. Assume 1 + 2 + … + k = k(k + 1)/2 for some k ≥ 1.This is an assumption, not a claim — say "assume", and name the k.
  3. Target. P(k+1) claims that the sum to k+1 equals (k + 1)(k + 2)/2.Substituting k+1 into the formula tells you exactly where the algebra must land.
  4. Sum to k+1 = [1 + 2 + … + k] + (k + 1) = k(k+1)/2 + (k + 1), using the hypothesis on the bracket.This is the only place the hypothesis is used, and using it must be visible.
  5. Factor out (k + 1): (k + 1)[k/2 + 1] = (k + 1)(k + 2)/2, which is the target.Taking out the common factor is almost always the move that finishes an induction on a sum.
  6. Conclusion. P(1) is true and P(k) ⟹ P(k+1), so by induction P(n) holds for all n ≥ 1. ∎The concluding sentence is a mark. Write it out every time.

Proved by induction for all n ≥ 1.

03

The binomial theorem

Expanding (a + b)² by hand is fine. (a + b)⁷ is not. The binomial theorem gives every term of the expansion directly, and its coefficients are the numbers in Pascal's triangle — which are exactly the ⁿCᵣ of the previous chapter, because choosing which brackets contribute a b is a combination problem.

Notice the pattern in the powers: they always total n. As the power of a falls from n to 0, the power of b rises from 0 to n, and there are n + 1 terms in all.

(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ, r = 0 … n(a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + bⁿgeneral term Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳthe (r+1)th term uses ⁿCᵣ — the index is one behind the term number, and that offset is where marks go
n
the power of the bracketa positive integer here
r
counts from 0so r = 0 gives the first term
ⁿCᵣ
the binomial coefficientrow n of Pascal’s triangle

Each entry is the sum of the two above it. Row n holds the coefficients of (a + b)n, and the same numbers answer "how many ways can r things be chosen from n" — the two chapters are one idea.

04

Finding one particular term

A full expansion is rarely wanted. The examinable skill is to reach into the middle of it and pull out one term — the term in x⁵, or the constant term, or the coefficient of x³ — without writing the rest.

The method: write the general term, simplify the power of x, set that power equal to what the question asks for, and solve for r. Then substitute that r back.

Worked example

Find the term independent of x in the expansion of (2x + 1/x²)⁹.

  1. General term: T_(r+1) = ⁹Cᵣ (2x)⁹⁻ʳ (1/x²)ʳ.Take a = 2x and b = x⁻², keeping the 2 inside the bracket where it belongs.
  2. Collect the powers of x: x⁹⁻ʳ × x⁻²ʳ = x⁹⁻³ʳ.Add the indices. This is the step that turns the question into an equation.
  3. "Independent of x" means the power is zero: 9 − 3r = 0, so r = 3.A constant term is the term in x⁰. If r had come out fractional, no such term would exist — a valid answer.
  4. Substitute r = 3: ⁹C₃ (2)⁶ = 84 × 64 = 5376.The 2⁶ comes from (2x)⁶ — forgetting to raise the coefficient 2 to its power is the usual error.

5376

Two facts worth carrying

Setting a = b = 1 in the theorem gives Σ ⁿCᵣ = 2ⁿ — the sum of any row of Pascal's triangle is a power of two, which also says a set of n elements has 2ⁿ subsets. And setting a = 1, b = −1 shows the alternating sum of a row is zero.

Before you leave this chapter

  1. Induction needs all four parts: basis, hypothesis, inductive step, conclusion.
  2. Write down what P(k+1) claims before starting the algebra — it gives the target.
  3. The hypothesis must be used visibly somewhere in the inductive step.
  4. (a + b)ⁿ has n + 1 terms; the powers of a and b always sum to n.
  5. T_(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ. To find a specific term, set the power of x equal to what is asked and solve for r.
06

The binomial series when the power is not a positive integer

The expansion of (1 + x)ⁿ stops after n + 1 terms only because ⁿCr eventually multiplies by zero. If n is negative or fractional that never happens, so the expansion continues forever — it becomes an infinite series.

An infinite series is only useful if it converges, and this one does so only when |x| < 1. That condition is not decoration: outside it the terms grow and the sum is meaningless. Stating the condition is almost always worth a mark on its own.

(1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + …valid for |x| < 1for (a + x)ⁿ: factor out a first(a + x)ⁿ = aⁿ(1 + x/a)ⁿ, valid for |x| < |a|the coefficients are written out in full, since ⁿCr has no meaning for fractional n
n
any rational numbernegative and fractional powers included
|x| < 1
the validity conditionwithout it the series does not converge
a
the constant termmust be factored out to reach the standard form
Worked example

Expand (4 + x)^(1/2) in ascending powers of x up to the term in x², and state the values of x for which the expansion is valid.

  1. Factor out the 4: (4 + x)^(1/2) = 4^(1/2)(1 + x/4)^(1/2) = 2(1 + x/4)^(1/2).The series only applies to (1 + something)ⁿ, so the constant must come out first. This step is where most marks are lost.
  2. Apply the series with n = ½ and x replaced by x/4: 1 + ½(x/4) + [½(−½)/2](x/4)².Every occurrence of x in the formula becomes x/4, including inside the square.
  3. = 1 + x/8 − (1/8)(x²/16) = 1 + x/8 − x²/128.The third coefficient is ½ × (−½) ÷ 2 = −1/8, and (x/4)² = x²/16.
  4. Multiply by the 2: 2 + x/4 − x²/64.The factored constant must be brought back in — forgetting it is the other standard error.
  5. Valid for |x/4| < 1, that is |x| < 4.The condition applies to whatever replaced x in the standard form, so it scales with the factored constant.

2 + x/4 − x²/64, valid for |x| < 4

The validity condition scales too

For (1 + 3x)ⁿ the condition is |3x| < 1, so |x| < ⅓. For (4 + x)ⁿ it is |x/4| < 1, so |x| < 4. Quoting |x| < 1 regardless of what was substituted is a guaranteed lost mark, and it is the most frequent error in this topic after forgetting to factor out the constant.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State the two steps of a proof by mathematical induction.
Model answer

Basis: show the statement is true for the first value, usually n = 1. Inductive step: assume it is true for n = k and prove it must then be true for n = k + 1. Together these prove it for every natural number.

Examiner tip. Both steps are needed, and the mark scheme wants the word "assume" in the second. A proof missing the basis proves nothing at all.

SQ2[2 marks]
How many terms are there in the expansion of (x + y)¹², and what is the coefficient of the third term?
Model answer

There are 12 + 1 = 13 terms. The third term uses r = 2, so its coefficient is ¹²C₂ = 66.

Examiner tip. The (r+1)th term uses ⁿCᵣ, so the third term uses r = 2, not r = 3. That offset is examined deliberately.

SQ3[2 marks]
Write down the general term in the expansion of (3x − 2)ⁿ.
Model answer

T_(r+1) = ⁿCᵣ (3x)ⁿ⁻ʳ (−2)ʳ.

Examiner tip. Keep the sign with the second term. Writing (2)ʳ and adding a minus later gets the sign wrong for every even r.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
Prove by mathematical induction that 2ⁿ > n for all natural numbers n.
Full working
  1. Basis: for n = 1, 2¹ = 2 > 1, so P(1) is trueboth sides evaluated[1]
  2. Assume 2ᵏ > k for some k ≥ 1the hypothesis stated as an assumption[1]
  3. Then 2ᵏ⁺¹ = 2 × 2ᵏ > 2k, using the hypothesisthe hypothesis is used here and must be visible[1]
  4. Since k ≥ 1, 2k = k + k ≥ k + 1, so 2ᵏ⁺¹ > k + 1; hence by induction the result holds for all nthe k ≥ 1 is what makes the last inequality work[1]

Proved by induction for all n ≥ 1.

Examiner tip. With inequalities the last step usually needs a small extra fact about k — here that k ≥ 1. Say where it comes from rather than asserting the inequality.

N2[4 marks]
Find the coefficient of x⁴ in the expansion of (2 + x)⁷.
Full working
  1. General term T_(r+1) = ⁷Cᵣ (2)⁷⁻ʳ xʳ[1]
  2. The power of x is r, so r = 4[1]
  3. ⁷C₄ = 35 and 2³ = 87 − 4 = 3, so the 2 is cubed[1]
  4. Coefficient = 35 × 8 = 280[1]

280

Examiner tip. The constant inside the bracket must be raised to its own power. Answering 35 — the binomial coefficient alone — is the standard incomplete answer.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
Consider the expansion of (x² − 2/x)⁶.
  1. Write down the general term.
  2. Find the term independent of x.
  3. State how many terms the full expansion has, and explain why not every expansion of this kind contains a constant term.
Mark scheme
  1. T_(r+1) = ⁶Cᵣ (x²)⁶⁻ʳ (−2/x)ʳthe minus stays with the 2[1]
  2. Powers of x: x¹²⁻²ʳ × x⁻ʳ = x¹²⁻³ʳ[1]
  3. Setting 12 − 3r = 0 gives r = 4[1]
  4. Term = ⁶C₄ (−2)⁴ = 15 × 16 = 240the power 4 is even, so the result is positive[1]
  5. The expansion has 6 + 1 = 7 terms[1]
  6. A constant term exists only if the equation for r has a solution that is a whole number between 0 and n; otherwise no term has power zero and there is no constant term[1]

(b) 240 (c) 7 terms; a constant term requires an integer solution for r

Examiner tip. Part (c) is the understanding mark. If solving for r gave 3.5, the honest answer would be "there is no term independent of x" — and that is sometimes exactly what is being asked.