数列{1/(4n^2-1)}的前n项和为Sn,求Sn

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数列{1/(4n^2-1)}的前n项和为Sn,求Sn
数列{1/(4n^2-1)}的前n项和为Sn,求Sn

数列{1/(4n^2-1)}的前n项和为Sn,求Sn
an=1/2(1/(2n-1)-1/(2n+1))
Sn=1/2(1-1/3+1/3-1/5+1/5-1/7+.+1/(2n-1)-1/(2n+1))=1/2(1-1/(2n+1)=n/(2n+1)