The first five numbers in a pattern of numbers are given in the table below - Leaving Cert Mathematics - Question 5 - 2016

To exhibit that the pattern is quadratic, we can find the first and second differences of the sequence:

1. First Differences:

   • U₂ - U₁ = 15 - 13 = 2
   • U₃ - U₂ = 19 - 15 = 4
   • U₄ - U₃ = 25 - 19 = 6
   • U₅ - U₄ = 33 - 25 = 8
   • U₆ - U₅ = 43 - 33 = 10
   • U₇ - U₆ = 55 - 43 = 12
   • U₈ - U₇ = 69 - 55 = 14

2. Second Differences:

   • 4 - 2 = 2
   • 6 - 4 = 2
   • 8 - 6 = 2
   • 10 - 8 = 2
   • 12 - 10 = 2
   • 14 - 12 = 2

The second differences are constant (equal to 2), which confirms that the pattern is quadratic.

Using the known terms:

1. For U₁:
   $1 + b + c = 13$
   This simplifies to:
   $b + c = 12$

2. For U₂:
   $4 + 2b + c = 15$
   This simplifies to:
   $2b + c = 11$

Solving these two equations:

• From the first equation, substitute $c = 12 - b$ into the second:

  $2b + (12 - b) = 11$
  $b + 12 = 11$
  $b = -1$

Substituting back to find c: $c = 12 - (-1) = 13$

Thus, the values are:

• b = -1
• c = 13

In the given arithmetic sequence:

• The first term, T₁ = 12
• The common difference, d = T₂ - T₁ = 14 - 12 = 2

Thus, the general term can be expressed as: $Tₙ = a + (n - 1)d$
Where $a = 12$ and $d = 2$. Therefore: $Tₙ = 12 + (n - 1) imes 2 = 12 + 2n - 2 = 10 + 2n$

To find T₃₀: $T₃₀ = 10 + 2(30) = 10 + 60 = 70$

Hence, the final answers are:

• Expression for Tₙ: $Tₙ = 10 + 2n$
• Value of T₃₀: 70