高中数学an=1/(n+1)(n+3)求sn
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αn=1/(n+1)(n+3)=1/2(1/(n+1)-1/(n+3)),sn=α1+α2+α3+……+αn=1/2(1/2-1/4)+1/2(1/2-1/5)+1/2(1/4-1/6)+……+1/2(1/(n+1)-1/(n+3))=1/2(1/2-1/4+1/3-1/5+1/4-1/6+……+1/(n+1)-1/(n+3))=1/2(1/2+1/3-1(1/n+3))=1/2(5/6-1/(n+3))
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2021-08-20 · 知道合伙人教育行家
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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)+(1/5-1/7)+(1/6-1/8)+(1/7-1/9)+……]
=(1/2)*[1/2+1/3-1/(n+2)-1/(n+3)]=5/12-(2n+5)/[2(n+2)(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)+(1/5-1/7)+(1/6-1/8)+(1/7-1/9)+……]
=(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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