记等差数列首项为a1,公差为d即an=a1+(n-1)d, n=1,2,....则an的平方和：Sn=∑(an)^2=∑[a1^2+(n-1)^2d^2+2a1d(n-1)]=na1^2+d^2∑(n-1)^2+a1d∑(n-1)=na1^2+d^2(n-1)n(2n-1)/6+a1d(n-1)n/2
