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+
b2
2
+
b3
3
+…+
bn
n
=an(n∈N*),求数列{bn}的通项公式;
(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,

2a1q=a1+a1q2-1

∵a1=1,∴2q=q2

∵q≠0,∴q=2,an=2n-1

b1+

b2
2
+
b3
3
+…+
bn
n
=an,①

当n≥2时,b1+

b2
2
+
b3
3
+…+
bn-1
n-1
=an-1,②

①-②,得

bn
n
=an-an-1=2n-1-2n-2=2n-2

∴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

=

1-2n-1
1-2
-n•2n-1

=(1-n)•2n-1-1,

∴Sn=(n-1)•2n-1+1.

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