数列{an}得通项公式为an=1/(4n-3)(4n+1)求sn

问题描述:

数列{an}得通项公式为an=1/(4n-3)(4n+1)求sn

1.将an转换
an = 1/[(4n-3)(4n+1)] ===>>an = 1/4[1/(4n-3)-1/(4n+1)]
即四分之一倍4n-3分之一减去4n+1分之一
4n-3分之一减去4n+1分之一是括号的内容
2.a1 = 1/5
3.sn = a1+a2+a3+……+an = 1/4[1-1/5+1/5-……+1/(4n-3)-1/(4n-1)}前后抵消后剩1/4[1-1/(4n+1)]=1/(4n+1)

An=1/(4n-3)(4n+1)=[1/(4n-3)-1/(4n+1)]/4
Sn=[1-1/5+1/5-1/9+...+1/(4n-7)-1/(4n-3)+1/(4n-3)-1/(4n+1)]/4
=[1-1/(4n+1)]/4
=n/(4n+1)