Mathematical Induction Calculator

Prove a sum formula by mathematical induction: the base case, the inductive step S(k) + T(k + 1) = S(k + 1) simplified exactly, and the first counterexample if the claim is false.

General term T(k)	k^2
Claimed sum S(n) = T(1) + … + T(n)	n(n + 1)(2n + 1)/6
Starting value of n	1

General term T(k)

The k-th term of the sum, in k. Example: k(k + 1), 1/(k(k + 1)), 2k − 1.

Claimed sum S(n) = T(1) + … + T(n)

The formula to prove, in n.

Starting value of n

The claim is

True for all n ≥ 1

P(1) is true, and P(k) ⇒ P(k + 1) because S(k) + T(k + 1) simplifies exactly to S(k + 1). By the principle of mathematical induction the formula holds for every n ≥ 1.

Exact fraction. Checked exactly.

Method: Mathematical induction
Worked out with: AllCalcHub exact polynomial algebra
Check: Every step recomputed with exact fractions; the inductive step as an identity of fractions of polynomials

P(1): 1 = 1
S(k + 1): (2k³ + 9k² + 13k + 6)/6

About the Mathematical Induction Calculator

To prove a statement P(n) for every n ≥ n₀, show P(n₀) is true (the base case), then show that P(k) true implies P(k + 1) true (the inductive step). For a sum formula, the step is S(k) + T(k + 1) = S(k + 1).

When the term and the formula are polynomials or fractions of polynomials, AllCalcHub checks the step as an exact identity, which is a complete proof.

The formula

P(n₀) and [P(k) ⇒ P(k + 1)] ⇒ P(n) for all n ≥ n₀

Worked example

Σk² = n(n + 1)(2n + 1)/6

n = 1: 1 = 1·2·3/6 ✓
k(k + 1)(2k + 1)/6 + (k + 1)² = (k + 1)(k + 2)(2k + 3)/6 ✓

Things to keep in mind

Divisibility statements (like 6 | 7ⁿ − 1) and inequalities aren't handled; for powers the formula is checked for 30 values rather than proved.

Questions people ask

Why isn't checking many values a proof?

A formula can work for the first million n and then fail. Induction covers every n with two finite checks.

What is strong induction?

Assuming P(n₀), …, P(k) all hold to prove P(k + 1). It is equivalent, and useful for recurrences.