Tile Pattern Analysis and Arithmetic Sequence Calculation

Table Data and Initial Observations

  • Figure 11 contains 55 tiles.

  • Figure 22 contains 99 tiles.

  • Figure 33 contains 1313 tiles.

  • Figure 44 contains 1717 tiles.

  • The sequence of tile counts corresponding to the sequential figure numbers is 55, 99, 1313, 1717, …\dots

Identification of Sequence Type and Common Difference

  • The difference between consecutive terms is evaluated as follows:

    • Difference between Figure 22 and Figure 11: 9−5=49 - 5 = 4

    • Difference between Figure 33 and Figure 22: 13−9=413 - 9 = 4

    • Difference between Figure 44 and Figure 33: 17−13=417 - 13 = 4

  • Because the difference between consecutive figure outputs remains constant (d=4d = 4), the system represents an arithmetic sequence (a linear pattern).

  • The initial value or first term is a1=5a_1 = 5.

  • The common difference between consecutive figures is d=4d = 4

Explicit Formula Derivation

  • The general formula for the nn\text{-th} term of an arithmetic sequence is given by: an=a1+(n−1)×da_n = a_1 + (n - 1) \times d

  • Substituting the established parameters (a1=5a_1 = 5 and d=4d = 4) into the standard formula gives: an=5+(n−1)×4a_n = 5 + (n - 1) \times 4

  • Distributing the common difference 44 across (n−1)(n - 1) yields: an=5+4×n−4a_n = 5 + 4 \times n - 4

  • Combining like terms simplifies the relation to the explicit linear equation: an=4×n+1a_n = 4 \times n + 1

Verification of the Explicit Formula

  • Verification for Figure 11 (n=1n = 1): a1=4×1+1=5a_1 = 4 \times 1 + 1 = 5

  • Verification for Figure 22 (n=2n = 2): a2=4×2+1=9a_2 = 4 \times 2 + 1 = 9

  • Verification for Figure 33 (n=3n = 3): a3=4×3+1=13a_3 = 4 \times 3 + 1 = 13

  • Verification for Figure 44 (n=4n = 4): a4=4×4+1=17a_4 = 4 \times 4 + 1 = 17

Calculation for Figure 10

  • Method 1: Explicit Formula Evaluation

    • To find the number of tiles in Figure 1010, substitute n=10n = 10 into the derived formula an=4×n+1a_n = 4 \times n + 1:   a10=4×10+1a_{10} = 4 \times 10 + 1   a10=40+1a_{10} = 40 + 1   a10=41a_{10} = 41

  • Method 2: Sequential Recursive Extension

    • Figure 11: 55\,\text{tiles}

    • Figure 22: 5+4=95 + 4 = 9\,\text{tiles}

    • Figure 33: 9+4=139 + 4 = 13\,\text{tiles}

    • Figure 44: 13+4=1713 + 4 = 17\,\text{tiles}

    • Figure 55: 17+4=2117 + 4 = 21\,\text{tiles}

    • Figure 66: 21+4=2521 + 4 = 25\,\text{tiles}

    • Figure 77: 25+4=2925 + 4 = 29\,\text{tiles}

    • Figure 88: 29+4=3329 + 4 = 33\,\text{tiles}

    • Figure 99: 33+4=3733 + 4 = 37\,\text{tiles}

    • Figure 1010: 37+4=4137 + 4 = 41\,\text{tiles}

  • Total tile count for Figure 1010 is 4141\,\text{tiles}.