已知等差数列an满足a3=7,a5+a7=26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/an^2-1(n属于N),求数列{bn}的前n项和Tn

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已知等差数列an满足a3=7,a5+a7=26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/an^2-1(n属于N),求数列{bn}的前n项和Tn

已知等差数列an满足a3=7,a5+a7=26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/an^2-1(n属于N),求数列{bn}的前n项和Tn
已知等差数列an满足a3=7,a5+a7=26
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/an^2-1(n属于N),求数列{bn}的前n项和Tn

已知等差数列an满足a3=7,a5+a7=26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/an^2-1(n属于N),求数列{bn}的前n项和Tn
a1+2d=7
2a1+10d=26
两方程联立得出a1和d
a1=3,d=2
公式an=a1+(n-1)d
sn=na1+[n(n-1)]d/2即可、

1、a1=3 d=2 所以an=2n+1 sn=n^2+n
2 bn=1/an^2-1(后面的1在分母上还是给整体减)

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