an = n²
Sn = 1² + 2² + 3² + ...... + n² = n(n+1)(2n+1)/6
归纳法证明:
n = 1, 1×(1+1)×(2×1+1)/6 = 6/6 = 1,求和公式正确
设 n = k 时，Sk = 1² + 2² + 3² + ...... + k² = k(k+1)(2k+1)/6 成立。
S(k+1) = k(k+1)(2k+1)/6+(k+1)²
= (k+1)[k(2k+1)/6+(k+1)]
= (k+1)[k(2k+1)+6k+6]/6
= (k+1)[2k²+7k+6]/6
= (k+1)[(k+2)(2k+3]/6
= (k+1)[(k+1)+1][2(k+1)+1]/6
得证。
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