Question

已知an=(2n-1)/2^(n-1),求和sn

Answer 1

简单计算一下即可，答案如图所示

分析

Answer 2

{(2n-1)/2^n}=
2n/2^n
-
1/2^n
对于后一部分
1/2^n
,
其前n项和为等比数列求和
S2
=
1/2
+
1/2^2
+
1/2^3
+
……
1/2^n
=
(1/2)
*
[1
-
(1/2)^n]/(1
-
1/2)
=
1
-
1/2^n
对于前一部分
2n/2^n
S1
=
2*(1/2
+
2/2^2
+
3/2^3
+
……
+
n/2^n)
两端乘2
2S1
=
2
*
[1
+
2/2
+
3/2^2
+
……
+
n/2^(n-1)]
两式相减,
将分母方次相同的项凑在一起
2S1
-
S1
=
S1
=
2*{
1
+
（2/2
-
1/2）+
(3/2^2
-
2/2^2)
+
……
+
[n/2^(n-1)
-
(n-1)/2^(n-1
)
-
n/2^n
}
=
2
*
[1
+
1/2
+
1/2^2
+
1/2^(n-1)
-
n/2^n]
=
2
*
{
1
*
[1
-
(1/2)^n]/(1
-1/2)
-
n/2^n}
=
2
*
[2
-
1/2^(n-1)
-
n/2^n]
=
4
-
4/2^n
-
2n/2^n
S
=
S1
-
S2
=
4
-
4/2^n
-
2n/2^n
-
1
+
1/2^n
=
3
-
(3
+
2n)/2^n