Question

1 patrat+2 patrat+3 patrat+...+n patrat=n(n+1)(2n+1) supra 6

Answer 1

se face prin inductie matematica

P(n) -> P(n+1)

P(n+1) = 1^2+2^2+3^2+...+n^2+(n+1)^2 = (n+1)(n+1+1)(2*(n+1)+1)/6

stim din ipoteza ca  1^2+2^2+3^2+...+n^2 = n(n+1)(2n+1)/6 si inlocuim in P(n+1)

P(n+1) = n(n+1)(2n+1)/6 + (n+1)^2 = (n+1)(n+1+1)(2*(n+1)+1)/6

(n^2+n)(2n+1)/6 + (n^2+2n+1) = (n+1)(n+2)(2n+3)/6

(2n^3+n^2+2n^2+n)/6 +n^2+2n + 1 = (n^2+2n+n+2)(2n+3)/6

(2n^3+3n^2+n)/6 + n^2+2n+1 = (n^2+3n+2)(2n+3)/6

(2n^3+3n^2+n)/6 + n^2+2n+1 = (2n^3+3n^2+6n^2+9n+4n+6)/6

(2n^3+3n^2+n)/6 + n^2+2n+1 = (2n^3+9n^2+13n+6)/6

2n^3+3n^2+n + 6n^2+12n+6 = 2n^3 + 9n^2 + 13n + 6

2n^3+9n^2+13n + 6 = 2n^3 + 9n^2 + 13n + 6

Adevarat!!!