Given a1 = 1, a2 = 2, and an+2 = 2an+1 − a_n + 3, find the 7th term of the sequence.

The first two terms are:
a₁ = 1
a₂ = 2
The recurrence relation is:
aₙ₊₂ = 2aₙ₊₁ - aₙ + 3

Substitute n = 1 into the recurrence relation:
a₁₊₂ = 2a₁₊₁ - a₁ + 3
a₃ = 2a₂ - a₁ + 3
a₃ = 2(2) - 1 + 3
a₃ = 4 - 1 + 3
a₃ = 6

Substitute n = 2 into the recurrence relation:
a₂₊₂ = 2a₂₊₁ - a₂ + 3
a₄ = 2a₃ - a₂ + 3
a₄ = 2(6) - 2 + 3
a₄ = 12 - 2 + 3
a₄ = 13

Substitute n = 3 into the recurrence relation:
a₃₊₂ = 2a₃₊₁ - a₃ + 3
a₅ = 2a₄ - a₃ + 3
a₅ = 2(13) - 6 + 3
a₅ = 26 - 6 + 3
a₅ = 23

Substitute n = 4 into the recurrence relation:
a₄₊₂ = 2a₄₊₁ - a₄ + 3
a₆ = 2a₅ - a₄ + 3
a₆ = 2(23) - 13 + 3
a₆ = 46 - 13 + 3
a₆ = 36

Substitute n = 5 into the recurrence relation:
a₅₊₂ = 2a₅₊₁ - a₅ + 3
a₇ = 2a₆ - a₅ + 3
a₇ = 2(36) - 23 + 3
a₇ = 72 - 23 + 3
a₇ = 52

The sequence of terms is 1, 2, 6, 13, 23, 36, 52.
The method shown in your notes, using a difference table, also confirms this result. The first differences are 1, 4, 7, 10, 13, 16. The second differences are constant at 3, which indicates a quadratic sequence.

The 7th term, a₇, is 52.