已知等差数列｛an｝满足:a3=7，a5+a7=26，｛an｝的前n项和为Sn.（

Question

已知等差数列｛an｝满足:a3=7，a5+a7=26，｛an｝的前n项和为Sn. （1）求an及Sn （2）令bn=1/a^n-1，求数列｛bn｝的前n项和Tn.

Answer 1

如果帮到您的话，可以好评吗？谢谢了！！！(右上角采纳)a3=7，a5+a7=26 a5+a7=26=2a6 a6=13 所以 d=(13-7)/(6-3)=2 an=a3+(n-3)d =7+2(n-3) =2n+1 bn=1/[(2n+1)²-1]=1/(2n+1-1)(2n+1+1)=1/[2n(2n+2)] Tn=b1+b2+....+bn =1/2×4＋1/4×6＋1/6×8＋....+1/[2n(2n+2)] =1/2 [1/2-1/4+1/4-1/6+1/6-1/8+....+1/2n-1/(2n+2)] =1/2 (1/2-1/(2n+2)) =1/2 × n/(2n+2) =n/(4n+4)