Let's make a table of differences
0 3 8 15 24
3 5 7 9
2 2 2
Constant second difference means a quadratic polynomial,
f(n)=an² + bn +c
f(1)=0, f(2)=3, f(3)=8
0 = a+b+c
3 = 4a + 2b + c
8 = 9a + 3b + c
Subtracting pairs
3 = 3a+b
5 = 5a + b
Subtracting,
2 = 2a
a = 1
b = 5 - 5a = 0
c = -a-b = -1
f(n)=n² -1
Duh, shoulda noticed that at the beginning.
f(17)=17²-1 = 288
a₁ =5
aₙ = aₙ₋₁ - 7
aₙ = a₁ + d(n-1)
aₙ = 5 - 7(n-1)
a₇ = 5 - 7(6) = -37
Answer: -37
