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The diagram shows dots and crosses in the first 3 patterns. (a) Complete the table. (b) Find (i) the number of crosses needed for pattern 45, (ii) the nth term for the sum of crosses and dots.

Mathematics

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2 parts
b(i)

137\text{137}

b(ii)

n2+3n+2n^2 + 3n + 2

Step-by-step solution

7 steps

Part a· 3 steps

Complete the table.

  1. Determine the pattern for the number of dots.

    The number of dots for pattern 1 is 1 (121^2).
    The number of dots for pattern 2 is 4 (222^2).
    This suggests the number of dots for pattern nn is n2n^2.
    For pattern 3, the number of dots is 32=93^2 = 9.

  2. Determine the pattern for the number of crosses.

    The number of crosses for pattern 1 is 5.
    The number of crosses for pattern 2 is 8.
    The difference between consecutive terms is 8−5=38 - 5 = 3. This is an arithmetic progression with a common difference of 3.
    For pattern 3, the number of crosses is 8+3=118 + 3 = 11.

  3. Calculate the sum of crosses and dots for pattern 3.

    For pattern 3, the number of dots is 9 and the number of crosses is 11.
    The sum of crosses and dots is 9+11=209 + 11 = 20.

    The completed table is:

    pattern123
    number of dots149
    number of crosses5811
    sum of crosses and dots61220

Part b(i)

137\text{137}

The number of crosses needed for pattern 45.

Find

  1. Find the $n$th term formula for the number of crosses.

    The sequence for the number of crosses is 5,8,11,…5, 8, 11, \dots.
    This is an arithmetic progression with the first term a=5a = 5 and the common difference d=8−5=3d = 8 - 5 = 3.
    The formula for the nnth term of an arithmetic progression is Tn=a+(n−1)dT_n = a + (n-1)d.
    Substitute a=5a=5 and d=3d=3:

    Tn=5+(n−1)3T_n = 5 + (n-1)3 Tn=5+3n−3T_n = 5 + 3n - 3 Tn=3n+2T_n = 3n + 2
  2. Calculate the number of crosses for pattern 45.

    Substitute n=45n=45 into the formula:

    T45=3(45)+2T_{45} = 3(45) + 2 T45=135+2T_{45} = 135 + 2 T45=137T_{45} = 137

    The number of crosses needed for pattern 45 is 137\boxed{\text{137}}.

Part b(ii)

n2+3n+2n^2 + 3n + 2

The $n$th term for the sum of crosses and dots.

  1. Analyze the sequence for the sum of crosses and dots.

    The sequence is 6,12,20,…6, 12, 20, \dots.
    Calculate the first differences:
    12−6=612 - 6 = 6
    20−12=820 - 12 = 8
    Calculate the second differences:
    8−6=28 - 6 = 2
    Since the second differences are constant, the nnth term is a quadratic sequence of the form Tn=an2+bn+cT_n = an^2 + bn + c.

  2. Determine the coefficients a,b,a, b, and cc.

    The second difference is equal to 2a2a:

    2a=2  ⟹  a=12a = 2 \implies a = 1

    The first term of the first differences is equal to 3a+b3a+b:

    3a+b=63a+b = 6

    Substitute a=1a=1:

    3(1)+b=63(1)+b = 6 3+b=6  ⟹  b=33+b = 6 \implies b = 3

    The first term of the sequence is equal to a+b+ca+b+c:

    a+b+c=6a+b+c = 6

    Substitute a=1a=1 and b=3b=3:

    1+3+c=61+3+c = 6 4+c=6  ⟹  c=24+c = 6 \implies c = 2

    Therefore, the nnth term for the sum of crosses and dots is Tn=1n2+3n+2T_n = 1n^2 + 3n + 2.
    The nnth term for the sum of crosses and dots is n2+3n+2\boxed{n^2 + 3n + 2}.

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