se face prin inductie matematica
P(n) implica P(n+1)
P(n+1) = 2^2+6^2+...+(4n-2)^2 + (4(n+1)-2)^2 = 4(n+1)(4(n+1)^2-1) / 3
P(n+1) =2^2+6^2+...+(4n-2)^2 +(4n+4-2)^2 = (4n+4)(4(n^2+2n+1)-1)/3
4n(4n^2-1)/3 + (4n+2)^2 = (4n+4)(4n^2+8n+4-1)/3
16n^3-4n/3 + 16n^2+16n+4 = (16n^3 +32n^2 + 12n + 16n^2 + 32n + 12)/3
16n^3-4n/3 + 16n^2+16n+4 = (16n^3 + 48n^2 + 44n + 12)/3
16n^3-4n + 48n^2 + 48n + 12 = 16n^3 + 48n^2 + 44n + 12
16n^3 + 48n^2 + 44n + 12 = 16n^3 + 48n^2 + 44n + 12
inductia este buna....