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Induction

Setting up recursions for tilings and restricted strings, then proving summation formulas and divisibility claims by mathematical induction rather than by counting.

Applied Combinatorics, Induction

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5 exercises

Exercise 1

A database uses record identifiers that are alphanumeric strings in which the 1010 decimal digits and 2626 upper-case letters are valid symbols. The criteria that define a valid record identifier are recursive. A valid record identifier of length n≥2n\geq 2 can be constructed in the following ways:

  • beginning with any upper-case letter other than DD and followed by any valid record identifier of length n−1n-1;
  • beginning…
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Exercise 3

Give a recursion for the number g(n)g(n) of ternary strings of length nn that do not contain 102102 as a substring.

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Exercise 4

A 2×n2\times n checkerboard is to be tiled using two types of tiles. The first tile is a 1×11\times 1 square tile. The second tile is called an LL-tile and is formed by removing the upper-right 1×11\times 1 square from a 2×22\times 2 tile. The LL-tiles can be used in any of the four ways they can be rotated. (That is, the “missing square” can be in any of four positions.) Let t(n)t(n) denote the number of tilings of the 2×n2\times n checkerboard using 1×11\times 1 tiles…

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Exercise 9

(A challenging problem) For each formula, give both a proof using the Principle of Mathematical Induction and a combinatorial proof. One of the two will be easier while the other will be more challenging.

  1. 12+22+32+⋯+n2=n(n+1)(2n+1)6\displaystyle 1^2+2^2+3^2+\dots+ n^2= \frac{n(n+1)(2n+1)}{6}

  2. (n0)20+(n1)21+(n2)22+⋯+(nn)2n=3n\displaystyle\binom{n}{0}2^0+\binom{n}{1}2^1+\binom{n}{2}2^2+\dots+\binom{n}{n}2^n=3^n

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Exercise 15

Use mathematical induction to prove that for all integers n≥1n\geq 1, is divisible by 99.

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Adapted from Applied Combinatorics © 2006–2025 Mitchel T. Keller, William T. Trotter. Licensed under CC BY-SA 4.0. Source: Applied Combinatorics. Imported at revision 07286c4.