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
(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+1,a1
∴(
-1)2=0,a1=1a1
当n=2时,2
=a2+1,1+a2
∴
=2,a2=3.a2+1
(II)∵2
=an+1,Sn
∴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