Question

已知数列{an}中,a1=5,an=2an-1=2^n-1(n∈N*且大于等于2),求数列{an}的前n项和Sn

Answer 1

是“an=2a(n-1)+2^n-1”吧？
an=2a(n-1)+2^n-1
an
-
1
=
2×[
a(n-1)
-
1
]
+2^n，
两边同时除以2^n，得
(an
-
1)/2^n
=
2×[
a(n-1)
-
1
]/2^n
+1
(an
-
1)/2^n
=
[
a(n-1)
-
1
]/2^(n-1)
+1
(an
-
1)/2^n
-
[
a(n-1)
-
1
]/2^(n-1)
=1，n≥2
即数列{(an
-
1)/2^n}是以(a1-1)/2=2为首项，1为公差的等差数列
∴(an
-
1)/2^n=2+(n-1)×1=n+1
an
-
1=(n+1)·2^n
令数列{an
-
1}的前n项和为Tn
则
Tn=
2×2
+
3×2²
+
4×2³
+
……
+
(n+1)·2^n
.........................①
2Tn=
2×2²
+
3×2³
+
……
+
n·2^n
+
(n+1)·2^(n+1)
..........................②
②－①，得
Tn=
(n+1)·2^(n+1)
-
(2²+2³+……+2^n)
-
4
=
(n+1)·2^(n+1)
-
(2+2²+2³+……+2^n)
-
2
=
(n+1)·2^(n+1)
-
[2(1-2^n)/(1-2)]
-
2
=
(n+1)·2^(n+1)
-
2^(n+1)
=
n·2^(n+1)
∴Sn=Tn
+
n=n·2^(n+1)
+
n