Suku ke n dari barisan 4,7,12,19 adalah

Ternyata beda tetap ditahap II, maka berlaku rumus Un = an^2 + bn + c
U1 = a (1)^2 + b(1) + c = 4===> a + b + c = 4
U2 = a (2)^2 + b(2) + c = 7 ===> 4a + 2b+c = 7
U3= a(3)^2 + b(3) + c = 12===> 9a + 3b + c = 12
a+b+c = 4
4a+2b+c = 7
_________ –
– 3a – b = -3

4a+2b+c = 7
9a+3b+c = 12
___________ –
– 5a – b = – 5

– 3a – b = – 3
– 5a – b = – 5
__________ –
2a = 2
a = 2/2
a = 1
– 3a – b = – 3 ===> – 3(1) – b = – 3
– 3 – b = – 3 ===> – b = – 3 + 3
– b = 0
b = 0
diperoleh a = 1, b = 0
a+b+c = 4
1 + 0 + c = 4
c = 4 – 1
c = 3
jadi: a = 1 ; b = 0 dan c = 3
maka rumus Un = an^2 + bn + c
Un = 1 (n)^2 + 0 (n) + 3
Un = n^2 + 3