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

1. Проверим для n=1: 1^2 = 1(1+1)(2*1+1)/6 = 1*2*3/6 = 1. Верно.

2. Предположим, верно для k: 1^2+...+k^2 = k(k+1)(2k+1)/6.

3. Проверим для k+1: 1^2+...+k^2+(k+1)^2 = k(k+1)(2k+1)/6 + (k+1)^2 = (k+1)[k(2k+1)/6 + (k+1)] = (k+1)[(2k^2+k+6k+6)/6] = (k+1)(2k^2+7k+6)/6 = (k+1)(k+2)(2k+3)/6. Доказано.