问题
解答题
已知在等比数列{an}中,2a2=a1+a3-1,a1=1. (1)若数列{bn}满足b1+
(2)求数列{bn}的前n项和Sn. |
答案
(1)设数列{an}的公比为q,2a2=a1+a3-1,
即2a1q=a1+a1q2-1,
∵a1=1,∴2q=q2,
∵q≠0,∴q=2,an=2n-1.
又b1+
+b2 2
+…+b3 3
=an,①bn n
当n≥2时,b1+
+b2 2
+…+b3 3
=an-1,②bn-1 n-1
①-②,得
=an-an-1=2n-1-2n-2=2n-2,bn n
∴bn=n•2n-2,n≥2.
∴bn=
.1,n=1 n•2n-1,n≥2
(2)由(1)得Sn=1+2×20+3×21+4×22+…+n•2n-2,③
2Sn=2+2×21+3×22+…+(n-1)•2n-2+n•2n-1,④
③-④得
-Sn=1+2+22+…+2n-2-n•2n-1
=
-n•2n-11-2n-1 1-2
=(1-n)•2n-1-1,
∴Sn=(n-1)•2n-1+1.