Question

已知数列{An}的前n项为Sn，且A1=1An+1=2Sn，（1）求A2，A3，A4的值 （2）求An

Answer 1

利用
an
=
sn
-
s<n-1>
sn
-
s<n-1>
=
2sn²/(2sn
-1)
(sn
-
s<n-1>)(2sn
-1)
=
2sn²
2sn²
-
sn
-
2sns<n-1>
+
s<n-1>
=
2sn²
sn
+
2sns<n-1>
-
s<n-1>
=
0
两端同时除以
sns<n-1>
1/sn
-
1/s<n-1>
=
2
所以
1/sn
是公差d=2的等差数列
1/sn
=
1/s1
+
(n-1)*d
1/sn
=
1/a1
+
2(n-1)
1/sn
=
1
+
2n
-
2
sn
=
1/(2n
-1)
n=1
时
a1
=
1
n≥2
时
an
=
sn
-
s<n-1>
=
1/(2n-1)
-
1/(2n-3)