All subjects · Further Mathematics · WFM01 Further Pure Mathematics 1

Proof by induction: series

Edexcel IAL Further Mathematics · WFM01 Further Pure Mathematics 1

Proof by induction shows a statement is true for every positive integer n. It works like a row of dominoes: show the first falls, and that each one knocks over the next.

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The structure

1. Basis: show the statement is true for n = 1.

2. Assumption: assume it is true for n = k, where k is a positive integer.

3. Inductive step: using the assumption, show it is true for n = k + 1.

4. Conclusion: "If the result is true for n = k, then it is true for n = k + 1. As it is true for n = 1, it is true for all positive integers n."

All four parts earn marks. Induction proves a given formula; it does not find one.

For a series

To prove Σ_(r=1)n f(r) = S(n):

Write down S(k + 1) first so you know your target. Take out common factors rather than expanding everything.

Example

Prove Σ r3 = n2(n + 1)2/4.

n = 1: LHS = 1, RHS = 1 × 4/4 = 1. True.

Assume Σ_(r=1)k r3 = k2(k + 1)2/4. Then

Σ_(r=1)k+1 r3 = k2(k + 1)2/4 + (k + 1)3 = (k + 1)2/4 × [k2 + 4(k + 1)] = (k + 1)2(k + 2)2/4,

which is the formula with n = k + 1. Then write the conclusion.

Worked example

Prove by induction that Σ_(r=1)n r2 = n(n + 1)(2n + 1)/6.

  1. n = 1: LHS = 1, RHS = 1 × 2 × 3/6 = 1. True for n = 1.
  2. Assume true for n = k: Σ_(r=1)k r2 = k(k + 1)(2k + 1)/6.
  3. Σ_(r=1)k+1 r2 = k(k + 1)(2k + 1)/6 + (k + 1)2 = (k + 1)/6 × [k(2k + 1) + 6(k + 1)] = (k + 1)(2k2 + 7k + 6)/6 = (k + 1)(k + 2)(2k + 3)/6.
  4. That is the formula with n = k + 1. True for n = 1, and true for n = k implies true for n = k + 1, so true for all positive integers n.

Answer: Proved for all n in ℤ+

Key idea: Basis, assumption, inductive step to the n = k + 1 form, conclusion. For series: S(k) + f(k + 1) = S(k + 1).

Check yourself

  1. What is the first step of a proof by induction?
    • Show the statement is true for n = 1
    • Assume the statement is true for n = k
    • Show the statement is true for n = k + 1
    • Check the statement for n = 1, 2 and 3 only
    Show answer

    Show the statement is true for n = 1. The basis case starts the chain.

  2. In the inductive step for Σ_(r=1)n f(r) = S(n), what do you add to S(k)?
    • f(k)
    • S(k + 1)
    • f(k + 1)
    • S(1)
    Show answer

    f(k + 1). Σ_(r=1)k+1 f(r) is the sum to k plus the next term, f(k + 1).

  3. To prove Σ_(r=1)n r3 = n2(n + 1)2/4, what expression should the inductive step reach?
    • (k + 1)2(k + 2)2/4
    • k2(k + 1)2/4
    • (k + 1)2 k2/4 + 1
    • (k + 1)3
    Show answer

    (k + 1)2(k + 2)2/4. Replace n with k + 1 in n2(n + 1)2/4.

  4. In the r3 proof, k2(k + 1)2/4 + (k + 1)3 = (k + 1)2/4 × [k2 + ak + 4]. Find a.
    Show answer

    4. (k + 1)3 = (k + 1)2 × 4(k + 1)/4, so the bracket is k2 + 4k + 4 = (k + 2)2.

  5. To prove Σ_(r=1)n r(r + 2) = n(n + 1)(2n + 7)/6, you check n = 1. What value do both sides give?
    Show answer

    3. LHS = 1 × 3 = 3. RHS = 1 × 2 × 9/6 = 3.

  6. Which is a suitable conclusion for an induction proof?
    • The formula works for n = 1, 2 and 3, so it is always true
    • True for n = k + 1, so true for all n
    • The formula is true for n = k
    • If true for n = k then true for n = k + 1; true for n = 1, so true for all positive integers n
    Show answer

    If true for n = k then true for n = k + 1; true for n = 1, so true for all positive integers n. The conclusion must link the basis to the inductive step and state the set of values.

  7. Use the r2 formula, proved by induction, to evaluate Σ_(r=1)12 r2.
    Show answer

    650. 12 × 13 × 25/6 = 650.

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Last updated 8 November 2026. Spotted a mistake? Tell us.