问题
解答题
设等差数列{an}的前n项和为Sn,且a2=4,S5=30.数列{bn}满足b1=0,bn=2bn-1+1,(n∈N,n≥2), ①求数列{an}的通项公式; ②设Cn=bn+1,求证:{Cn}是等比数列,且{bn}的通项公式; ③设数列{dn}满足dn=
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答案
①由a2=a1+d=4,S5=5a1+
d=30得:a1=2,d=2,5×4 2
∴an=2+2(n-1)=2n…(4分)
②∵bn=2bn-1+1,cn=bn+1,
∴
=cn cn-1
=bn+1 bn-1+1
=2(n≥2,n∈N)2(bn-1+1) bn-1+1
∴{cn}是以2为公比的等比数列.
又∵c1=b1+1=1,
∴cn=bn+1=1×2n-1=2n-1,
∴bn=2n-1-1…(9分)
③∵dn=
+bn=4 an•an+1
+2n-1-1=(4 2n•2(n+1)
-1 n
)+2n-1-1,1 n+1
∴Tn=[(1-
)+(1 2
-1 2
)+…+(1 3
-1 n
)]+(1+2+22+…+2n-1)-n1 n+1
=(1-
)+1 n+1
-n1-2n 1-2
=2n-n-
(14分)1 n+1