Question

2^2+6^2+...+(4n-2)^2=4n(4n^2-1) supra 3. Va rog, grabiti-va!!!

Answer 1

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....