Buktikan bahwa2 + 4 + 6 + 8 +… +2n = n kuadrat + n

Deret aritmatika
2, 4, 6, 8, …., 2n
suku pertama = a = 2
selisih = U2-U1 = 4-2 = 2 = b

jumlah deret aritmatika
Sn = n/2 (2a+(n-1)b)
= n/2 (2.2+(n-1)2)
= n/2. (4+2n-2)
= n/2. (2n+2)
= n²+n
(terbukti)