题目

若Cn=1/[(4n-3)(4n+1)],求数列{cn}的前n项和Tn

提问时间：2021-03-29

答案

Cn=1/4*4/[(4n-3)(4n+1)]
=1/4*[(4n+1)-(4n-3)]/[(4n-3)(4n+1)]
=1/4*[(4n+1)/[(4n-3)(4n+1)]-(4n-3)/[(4n-3)(4n+1)]]
=1/4[1/(4n-3)-1/(4n+1)]
所以Tn=1/4*[1-1/5+1/5-1/9+……+1/(4n-3)-1/(4n+1)]
=1/4*[1-1/(4n+1)]
=n/(4n+1)