求和：Sn=1*n+2*(n-1)+3*(n-2)+……+n*1

求和：Sn=1*n+2*(n-1)+3*(n-2)+……+n*1
求和：Sn=1*n+2*(n-1)+3*(n-2)+……+n*1

求和：Sn=1*n+2*(n-1)+3*(n-2)+……+n*1
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Sn+1=1*(n+1)+2*(n)+3*(n-1)+……+(n+1)*1
=1*n+1 + 2*(n-1)+2 + 3*(n-2)+3 +……+ n*1+n
=1*n+2*(n-1)+3*(n-2)+……+n*1 + 1+2+...+n
=Sn + 1/2n(n+1)=1/2n²+1/2n
累加得 Sn+1=(Sn+1 - Sn)+(Sn - Sn-1)+...+(S2-S1)+S1
=1/2(1²+2²+...+n²)+1/2(1+2+...+n)+1
=1/12n(n+1)(2n+1)+1/4n(n+1)+1
故Sn=1/12n(n-1)(2n-1)+1/4n(n-1)+1=1/6(n-1)n(n+1)+1 (n∈N*)

Sn=n*1+(n-1)*2+(n-2)*3+...+2*(n-1)+1*n(n∈N*) 共N项
Sn+1=(n+1)*1+(n)*2+(n-1)*3+...3*(n-1)+2*(n)+1*(n+1)(n∈N*) 共N+1项
两式相减
an+1=Sn+1-Sn=1*1+1*2+1*3+...+1*(n-1)+1*n+1*(n+1)
=1+2+3+4+.......

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Sn=n*1+(n-1)*2+(n-2)*3+...+2*(n-1)+1*n(n∈N*) 共N项
Sn+1=(n+1)*1+(n)*2+(n-1)*3+...3*(n-1)+2*(n)+1*(n+1)(n∈N*) 共N+1项
两式相减
an+1=Sn+1-Sn=1*1+1*2+1*3+...+1*(n-1)+1*n+1*(n+1)
=1+2+3+4+....+(n+1)=(n+1)(n+2)/2
an=n(n+1)/2=(n^2+n)/2
Sn求和时将括号内两项分别求和
n^2 项求和为 1/6n(n+1)(2n+1)
n 项求和为 n(n+1)/2
所以Sn=[1/6n(n+1)(2n+1)+n(n+1)/2]/2 =n(n+1)(n+2)/6

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