Question

已知数列an满足a1=1,an-2an-1-2^(n-1)=0(n属于正整数，n≥2，设bn=an/2^n

求b1,b2的值，并证明数列bn是等差数列 求数列an的前n项和Sn

Answer 1

^^an
-2a(n-1)
-2^(n-1)=0
an/2^n
-
a(n-1)/2^(n-1)
=
1/2
bn=an/2^n
是
等差数列
,
d=1/2
an/2^n
-
a1/2^1
=
(n-1)/2
an/2^n
=
n/2
an
=
n.2^(n-1)
b1=
1/2
b2=
1
let
S
=
1.2^0+2.2^1+....+n.2^(n-1)
(1)
2S
=
1.2^1+2.2^2+....+n.2^n
(2)
(2)-(1)
S
=
n.2^n
-
[1+2+...+2^(n-1)
]
=n.2^n
-
(2^n-1)
=
1
+
(n-1).2^n
Sn
=
a1+a2+...+an
=
S
=
1
+
(n-1).2^n