Question

已知等差数列｛an｝中，a2+a8＝26,且a3＝7 1求数列｛an｝的通向公式 2设bn＝1/a

已知等差数列｛an｝中，a2+a8＝26,且a3＝7
1求数列｛an｝的通向公式
2设bn＝1/anan+1,求数列｛bn｝的前n项和Sn

Answer 1

(1)
an = a1+(n-1)d

a2+a8=26
2a1+8d=26
a1+4d =13 (1)

a3=7
a1+2d = 7 (2)

(1)-(2)
d = 3

from (1) =>a1=1
an = 1+3(n-1) = 3n -2
(2)
bn＝1/(an.a(n+1)),求数列｛bn｝的前n项和Sn

bn＝1/(an.a(n+1))
= 1/[(3n-2)(3n+1)]
=(1/3) [ 1/(3n-2)-1/(3n+1) ]
Sn =b1+b2+...+bn
=(1/3) [ 1 -1/(3n+1) ]
=n/(3n+1)

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真的？

追答

应该没错！

追问

第一个小题跟第二个小题从哪分开的，怎么分不清楚

追答

(1)
an = a1+(n-1)d

a2+a8=26
2a1+8d=26
a1+4d =13 (1)

a3=7
a1+2d = 7 (2)

(1)-(2)
d = 3

from (1) =>a1=1
an = 1+3(n-1) = 3n -2
-----------------------------
(2)
bn＝1/(an.a(n+1)),求数列｛bn｝的前n项和Sn

bn＝1/(an.a(n+1))
= 1/[(3n-2)(3n+1)]
=(1/3) [ 1/(3n-2)-1/(3n+1) ]
Sn =b1+b2+...+bn
=(1/3) [ 1 -1/(3n+1) ]
=n/(3n+1)

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