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已知数列an=1/[n(n+1)(n+2)]求和sn(即求a1+a2+a3+a4+...+an)
题目详情
已知数列an=1/[n(n+1)(n+2)]
求和sn
(即求a1+a2+a3+a4+...+an)
求和sn
(即求a1+a2+a3+a4+...+an)
▼优质解答
答案和解析
an=1/[n(n+1)(n+2)]=1/2[1/n(n+1)-1/(n+1)(n+2)]
Sn=1/1*2*3+1/2*3*4+.+1/[n(n+1)(n+2)]
=1/2[1/1*2-1/2*3]+1/2[1/2*3-1/3*4]+...+1/2[1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/2*3+1/2*3-1/3*4+...+1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/(n+1)(n+2)]
=1/2*[(n+1)(n+2)-2]/[2(n+1)(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]
Sn=1/1*2*3+1/2*3*4+.+1/[n(n+1)(n+2)]
=1/2[1/1*2-1/2*3]+1/2[1/2*3-1/3*4]+...+1/2[1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/2*3+1/2*3-1/3*4+...+1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/1*2-1/(n+1)(n+2)]
=1/2*[(n+1)(n+2)-2]/[2(n+1)(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]
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