InduKSi Matematika
i. n= 1
ii. n = k
iii. n = k+1

Penjelasan dengan langkah-langkah:

untuk setiap x  ∈ A , berlaku

4+ 10+ 16 + . . . + (6n -2) =  n(3n +1)

__

i. n = 1
6(1) - 2  =  1(3(1)+ 1)
4 = 1 (4)
4 = 4

ii. n = k

4+ 10+ 16 + . . . + (6k -2) =  k(3k +1)

iii. n =k + 1

4+ 10+ 16 + . . . + (6k -2) +{ 6(k+1) - 2} =  (k+1) {3(k +1) + 1 }

k(3k + 1) + 6(k+1) - 2 =  (k+1)(3k+ 3 + 1)
3k²  + k + 6k + 6 - 2 = (k+ 1)(3k+4)

3k² + 7k + 3 = (k+ 1)(3k+4)
(k+ 1)(3k+4) = (k+ 1)(3k+4)

ruas kiri =  ruas kanan
terbukti