请大家看一下，这道题怎么做？ 1／2×3×4×5＋1／3×4×5×6＋...＋1／12×13×14×15＋1／13×14×15×1
1／2×3×4×5＋1／3×4×5×6＋...＋1／12×13×14×15＋1／13×14×15×16...
1／2×3×4×5＋1／3×4×5×6＋...＋1／12×13×14×15＋1／13×14×15×16
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利用 1/[(n+1)(n+2)]=1/(n+1)-1/(n+2),
通项 An=1/[(n+1)(n+2)(n+3)(n+4)]=[1/(n+1)-1/(n+2)]*[1/(n+3)-1/(n+4)]
=1/[(n+1)(n+3)]-1/[(n+2)(n+3)]-1/[(n+1)(n+4)]+1/[(n+2)(n+4)]
=[1/(n+1)-1/(n+3)]/2-[1/(n+2)-1/(n+3)]-[1/(n+1)-1/(n+4)]/3+[1/(n+2)-1/(n+4)]/2
=[1/(n+1)-1/(n+4)]/6-[1/(n+2)-1/(n+3)]/2,
求和 S(13)={(1/2-1/5)/6-(1/3-1/4)/2}+{(1/3-1/6)/6-(1/4-1/5)/2}+……+{(1/13-1/16)/6-(1/14-1/15)/2}+{(1/14-1/17)/6-(1/15-1/16)/2}
=[(1/2-1/5)+(1/3-1/6)+……+(1/13-1/16)+(1/14-1/17)]/6-[(1/3-1/4)+(1/4-1/5)+……+(1/14-1/15)+(1/15-1/16)]/2
=[1/2+1/3+1/4-1/15-1/16-1/17]/6-[1/3-1/16]/2
=169/12240 .
通项 An=1/[(n+1)(n+2)(n+3)(n+4)]=[1/(n+1)-1/(n+2)]*[1/(n+3)-1/(n+4)]
=1/[(n+1)(n+3)]-1/[(n+2)(n+3)]-1/[(n+1)(n+4)]+1/[(n+2)(n+4)]
=[1/(n+1)-1/(n+3)]/2-[1/(n+2)-1/(n+3)]-[1/(n+1)-1/(n+4)]/3+[1/(n+2)-1/(n+4)]/2
=[1/(n+1)-1/(n+4)]/6-[1/(n+2)-1/(n+3)]/2,
求和 S(13)={(1/2-1/5)/6-(1/3-1/4)/2}+{(1/3-1/6)/6-(1/4-1/5)/2}+……+{(1/13-1/16)/6-(1/14-1/15)/2}+{(1/14-1/17)/6-(1/15-1/16)/2}
=[(1/2-1/5)+(1/3-1/6)+……+(1/13-1/16)+(1/14-1/17)]/6-[(1/3-1/4)+(1/4-1/5)+……+(1/14-1/15)+(1/15-1/16)]/2
=[1/2+1/3+1/4-1/15-1/16-1/17]/6-[1/3-1/16]/2
=169/12240 .
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因为1/n(n+1)(n+2)(n+3)=1/2(1/n(n+3)-1/n(n+2))=1/2(1/3(1/n-1/n+3)-(1/n+1-1/n+2)) 所以原式等于:1/6(1/2-1/5+1/3-1/6+......+1/13-1/16)+1/2(1/3-1/4+1/4-1/5+......+1/12-1/13)=1/6(1/2+1/3+1/4-1/14-1/15-1/16)+1/2(1/3-1/13)=......(最终结果自己算一下吧)
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