Suku pertama deret aritmatika Sn=4n²—n. Tentukan suku kelima deret aritmatika

S1 = 4 X 1^2 - 1
= 4 - 1
=3
S1 = U1
Un = 8n - 5
U5 = 8. 5 - 5
= 40 -5
= 35

5n - 4n² + n = 0
6n - 4n² = 0
2n x (3 - 2n) = 0
n x (3 - 2n) = 0
n = 0
3 - 2n = 0
n = 0
n = 3|2 jadi, hasil akhirnya n1 = 0, n² = 3|2