,a
1a
2∴c
12,即c
2(
≥2),即当
≥2时,b
2
1,又当
1时,b12a12
∴b
于是S
b1b2b3…b
22324…2
12
26,
≥2,
.
11.解1因为S1=a1,S2=2a1+2×21×2=2a1+2,S4=4a1+4×23×2=4a1+12,由题意得2a1+22=a14a1+12,解得a1=1,所以a
=2
-12b
=-1
-1a
4a
+1=-1
-12
-14
2
+1=-1
-12
1-1+2
1+1.当
为偶数时,T
=1+13-13+15+…+2
1-3+2
1-1-2
1-1+2
1+1=1-2
1+1=2
2+
1当
为奇数时,T
=1+13-13+15+…-2
1-3+2
1-1+2
1-1+2
1+1=1+2
1+1=22
++21
f2
+2
,
为奇数,2
+1所以T
=2
2+
1,
为偶数
2
+1+-1
-1
或T
=
2
+1
12.1解由S2
-
2+
-1S
-
2+
=0,得S
-
2+
S
+1=0,由于a
是正项数列,所以S
+10所以S
=
2+
∈N.
≥2时,a
=S
-S
-1=2
,
=1时,a1=S1=2适合上式.∴a
=2
∈N.
2证明由a
=2
∈N得b
=
+
+212a2
=4
2
+
+122=116
12-
+122T
=1161-312+212-412+312-512+…+
-112-
+112+
12-
+122=1161+212-
+112-
+1221161+212=654
∈N.即对于任意的
∈N,都有T
654
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