问题 解答题
设{an}是正数组成的数列,其前n项和为Sn,且对于所有的正整数n,有2
Sn
=an+1

(I)求a1,a2的值;
(II)求数列{an}的通项公式;
(III)令b1=1,b2k=a2k-1+(-1)k,b2k+1=a2k+3k(k=1,2,3,…),求数列{bn}的前2n+1项和T2n+1
答案

(I)当n=1时,2

a1
=a1+1,

(

a1
-1)2=0,a1=1

当n=2时,2

1+a2
=a2+1,

a2+1
=2,a2=3.

(II)∵2

Sn
=an+1,

∴4Sn=(an+1)24Sn-1=(an-1+1)2,相减得:(an+an-1)(an-an-1-2)=0

∵{an}是正数组成的数列,

∴an-an-1=2,∴an=2n-1.

(Ⅲ)T2n+1=b1+[a1+(-1)1]+(a2+31)+[a3+(-1)2]+(a4+32)++(a2n+3n

=1+S2n+(3+32++3n)+[(-1)1+(-1)2++(-1)n]

=1+(2n)2+

3(1-3n)
1-3
+
(-1)(1-(-1)n)
1-(-1)

=

3n+1-2+8n2+(-1)n
2

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