n(n+1)=n^2+n
æ以Sn=(1^2+1)+(2^2+2)+...+(n^2+n)
=(1^2+2^2+...+n^2)+(1+2+...+n)
=n(n+1)(2n+1)/6+n(n+1)/2
=n(n+1)(n+2)/3
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追é®n(n+1)(2n+1)/6+n(n+1)/2è¿ä¸æ¥æä¹æ¥çï¼
追ç1+2+...+n=n(n+1)/2ç¥éå§
1^2+2^2+...+n^2=n(n+1)(2n+1)/6
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