Question

已知等差数列｛an｝的前n项和为Sn，a5=5，S5=15，则数列｛1/(an*a（n+1）)｝的前一百项和为多少？

Answer 1

∵｛an｝是等差数列
∴S5=5(a1+a5)/2
=5(a1+5)/2
=15
解得：a1=1
则d=(a5-a1)/4=1
∴通项公式：an=a1+(n-1)d=n
则数列｛1/(an*a（n+1）)｝的前一百项和为：
1/(1x2)+1/(2x3)+……+1/(100x101)
=(1-1/2)+(1/2-1/3)+……+(1/100-1/101)
=1-1/2+1/2-1/3+……+1/100-1/101
=1-1/101
=100/101

Answer 2

回答 共1条
检举|今天 09:37wdxf4444|十二级∵｛an｝是等差数列
∴S5=5(a1+a5)/2
=5(a1+5)/2
=15
解得：a1=1
则d=(a5-a1)/4=1
∴通项公式：an=a1+(n-1)d=n
则数列｛1/(an*a（n+1）)｝的前一百项和为：
1/(1x2)+1/(2x3)+……+1/(100x101)
=(1-1/2)+(1/2-1/3)+……+(1/100-1/101)
=1-1/2+1/2-1/3+……+1/100-1/101
=1-1/101
=100/101