已知s1=1,s2=1+2,s3=1+2+3,.sn=1+2+3+.+n,求Dn=s1+s2+s3,

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因为sn=n^2/2+n/2
所以dn=(1^2/2+1/2)+(2^2/2+2/2)+(3^2/2+3/2)+...+(n^2/2+n/2)
=1/2(1^2+2^2+3^2+...+n^2)+1/2(1+2+3+...+n)
=1/2*n/6(2n+1)(n+1)+1/2*n*(n+1)/2
=(2n+1)(n+1)/12+n(n+1)/4
======以下答案可供参考======
供参考答案1:
Dn=(1^2+2^2+3^2+…+n^2)/2+(1+2+…+n)/2
=n*(n+1)*(2*n+1)/12+n*(n+1)/4
=n*(n+1)*(n+2)/6