已知数列{an}的通项公式为an=1/(n+1)(n+3),求数列{an}的前n项和Sn
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an=1/[(n+1)(n+3)]=(1/2)[1/(n+1)-1/(n+3)]
Sn=a1+a2+...+an
=(1/2)[1/2-1/4+1/3-1/5+...+1/(n+1)-1/(n+3)]
=(1/2)[ [1/2+1/3+...+1/(n+1)]-[1/4+1/5+...+1/(n+1)+1/(n+2)+1/(n+3)] ]
=(1/2)[1/2+1/3-1/(n+1)-1/(n+2)]
=(5/12) -1/[2(n+1)]-1/[2(n+2)]
Sn=a1+a2+...+an
=(1/2)[1/2-1/4+1/3-1/5+...+1/(n+1)-1/(n+3)]
=(1/2)[ [1/2+1/3+...+1/(n+1)]-[1/4+1/5+...+1/(n+1)+1/(n+2)+1/(n+3)] ]
=(1/2)[1/2+1/3-1/(n+1)-1/(n+2)]
=(5/12) -1/[2(n+1)]-1/[2(n+2)]
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an=1/[(n+1)(n+3)]=(1/2)[1/(n+1)-1/(n+3)]
Sn=(1/2)[1/2-1/4]+[1/3-1/5]+[1/4-1/6]+…+[1/n-1/(n+2)]+[1/(n+1)-1/(n+3)]
=(1/2)[1/2+1/3--1/(n+2)-1/(n+3)]
=5/12-(2n+5)/[2(n+2)(n+3)]
Sn=(1/2)[1/2-1/4]+[1/3-1/5]+[1/4-1/6]+…+[1/n-1/(n+2)]+[1/(n+1)-1/(n+3)]
=(1/2)[1/2+1/3--1/(n+2)-1/(n+3)]
=5/12-(2n+5)/[2(n+2)(n+3)]
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列项求和
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