己知等差数列（an）满足：a3=7,a5+a7=26,(an)前n项和为S（

己知等差数列（an）满足：a3=7,a5+a7=26,(an)前n项和为S（

解：
设该等差数列的公差为d,则：
∵a5+a7=a3+a9=26
∴a9=26-7=19=a3+6d=7+6d
∴d=2
a3=a1+2d=7
a1=3
∴an=a1+(n-1)d
=3+2(n-1)
=2n+1
Sn=(a1+an)/2 = (3+2n+1)n/2 = n(n+2)
bn=1/Sn=1/[n(n+2)]=(1/2)[1/n - 1/(n+2)]
b1=(1/2)(1-1/3)
b2=(1/2)(1/2-1/4)
b3=(1/2)(1/3-1/5)
b4=(1/2)(1/4-1/6)
......
bn=(1/2)[1/n - 1/(n+2)]
上述各式相加：
Tn=(1/2)[1+1/2 -1/(n+1) - 1/(n+2)]=3/4-(n+3/2)/[(n+1)(n+2)]

(1)a5+a7=2a3+2d+4d=26,解得d=2 故a1=3 an=2n+1
sn=n（n+2）
（2）bn=1/2((1/n)-1/(n+2))
tn=1/2(3/2-((2n+3)/(n+1)(n+2)))

(1)a5+a7=2a6=26,a6=13
所以d=(a6-a3)/3=2
a1=a3-2d=3
所以an=2n+1
Sn=(a1+an)*n/2=n^2+2n
(2)bn=1/Sn=1/[n(n+2)]=1/2[1/n-1/(n+1]
所以Tn=1/2(1-1/3)+1/2(1/2-1/4)+……+1/2(1/n-1/(n+2))=1/2[1+1/2-1/(n+1)-1/(n+2)]
=1/2[3/2-(2n+3)/(n+1)(n+2)]

a5+a7=2a3+6d=14+6d=26，d=2
a1=a3-2d=7-2×2=3
an=3+2（n-1）=2n+1
Sn=（a1+an）×n/2=n（n+2）
2、b1=1/1×3，b2=1/2×4，……，bn=1/n（n+2）
Tn=1/1×3+1/2×4+……+1/n（n+2）=1/2×（1-1/3+1/2-1/4+……+1/n-1/n+2）
=3n²+5n/4（n+1）（n+2）

差数列，则有a5+a7=2a6=26，得到a6=13。a6-a3=3d=13-7=6，得到公差d=2。.
可得a3-2d=a1=7-4=3。.an=3+2(n-1)=2n+1.
Sn=na1+n(n-1)d/2=3n+n(n-1)=n^2+2n。
bn=1/Sn=1/n(n+2)=1/2[1/n-1/(n+2)]
Tn=1/2[1-1/3]+1/2[1/2-1/4]+1/2[1/3-1/5]+...+1/2[1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
=3/4-1/2*(2n+3)/(n+1)(n+2)