a1=1,an+1=2an+2^n 设bn=an/2^n-1 1证明bn是等差数列 2求an前n项和sna1=1,an+1=2an+2^n 设bn=an/2^n-1 1.证明bn是等差数列 2.求an前n项和sn

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a1=1,an+1=2an+2^n 设bn=an/2^n-1 1证明bn是等差数列 2求an前n项和sna1=1,an+1=2an+2^n 设bn=an/2^n-1 1.证明bn是等差数列 2.求an前n项和sn

a1=1,an+1=2an+2^n 设bn=an/2^n-1 1证明bn是等差数列 2求an前n项和sna1=1,an+1=2an+2^n 设bn=an/2^n-1 1.证明bn是等差数列 2.求an前n项和sn
a1=1,an+1=2an+2^n 设bn=an/2^n-1 1证明bn是等差数列 2求an前n项和sn
a1=1,an+1=2an+2^n 设bn=an/2^n-1 1.证明bn是等差数列 2.求an前n项和sn

a1=1,an+1=2an+2^n 设bn=an/2^n-1 1证明bn是等差数列 2求an前n项和sna1=1,an+1=2an+2^n 设bn=an/2^n-1 1.证明bn是等差数列 2.求an前n项和sn
bn = an/2^(n-1)
b = a/2^(n-2)
bn - b
= an/2^(n-1) - a/2^(n-2)
= (an - 2a )/2^(n-1)
把 已知条件 a = 2an+2^n 即 an = 2a + 2^(n-1) 代入上式
bn - b
= 2^(n-1)/2^(n-1)
= 1
因此 bn 是等差数列
b1 = a1/2^(1-1) = 1/1 = 1
bn = n
--------------------
an/2^(n-1) = n
所以
an = n * 2^(n-1)
-------------------------
Sn = a1 + a2 + a3 + …… + a + an
= 1 + 2*2 + 3*2^2 + …… + (n-1)*2^(n-2) + n * 2^(n-1)
2Sn = 2 + 2*2^2 + 3*2^3 + …… + (n-1)*2^(n-1) + n * 2^n
两式子相减, 把 2的乘方相同的相合并在一起
2Sn - Sn = Sn
= -1 + (1-2)*2 + (2-3)*2^2 + (3-4)*2^3 + …… [(n-1) -n]*2^(n-1) + n*2^n
= n*2^n - [ 1 + 2 + 2^2 + …… 2^(n-1)]
= n*2^n - 1*(2^n -1)/(2-1)
= n * 2^n - 2^n + 1
= (n-1)*2^n + 1

解: (1)
a(n+1)=2an+2^n
a(n+1)/2^n=2an/2^n+1
a(n+1)/2^n=an/2^(n-1)+1
由于 bn=an/2^(n-1)
则b(n+1)=a(n+1)/2^n
则: b(n+1)=bn+1
b(n+1)-bn=1
则{bn}是等差数列
(2...

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解: (1)
a(n+1)=2an+2^n
a(n+1)/2^n=2an/2^n+1
a(n+1)/2^n=an/2^(n-1)+1
由于 bn=an/2^(n-1)
则b(n+1)=a(n+1)/2^n
则: b(n+1)=bn+1
b(n+1)-bn=1
则{bn}是等差数列
(2)由于b1=a1/2^0
=1
则: bn=b1+(n-1)*1
=n
则: an/2^(n-1)=n
an=n*2^(n-1)
则:Sn=a1+a2+a3+...+an
=1*2^0+2*2^1+3*2^2+...+n*2^(n-1)
=(n+1)*2^n-1

收起

a1 = 1
S1 = a1 = 1
以 n = 1 代入一楼计算结果， S1 = (1+1)*2^1 -1 = 3 , 错误！
以 n = 1 2 3 代入二楼结果， 正确！

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