Question

己知等差数列（an）满足：a3=7,a5+a7=26,(an)前n项和为S（1）求an及Sn（2）令bn=1/Sn(n属

Answer 1

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)]