a1=1 an+1=2an+2^(n+1) 求通项公式

A(n+1)=2An+2^n
两边除以2^(n+1)
A(n+1)/2^(n+1)=2An/2^(n+1)+2^n/2^(n+1)=An/2^n+1/2
{An/2^n}是公差为1/2的等差数列
A1/2^1=1/2
An/2^n=n/2
An=n×2^(n-1)
Sn=A1+A2+A3+……+An
=1×2^0+2×2^1+3×2^2+……+n×2^(n-1)
2Sn=1×2^1+2×2^2+3×2^3+……+n×2^n
两式错位相减
Sn-2Sn=1+[(2×2^1-1×2^1)+(3×2^2-2×2^2)+……+n×2^(n-1)-(n-1)×2^(n-1)]-n×2^n
=1+(2^1+2^2+……+2^(n-1))-n×2^n
=1×(1-2^n)/(1-2)-n×2^n
=(1-n)×2^n-1
Sn=1+(n-1)×2^n

(1)
a(n+1)=2an+2^(n+1)
等式两边同除以2^(n+1)
a(n+1)/2^(n+1)=an/2ⁿ+1
a(n+1)/2^(n+1)-an/2ⁿ=1,为定值
a1/2=1/2,数列{an/2ⁿ}是以1/2为首项,1为公差的等差数列
bn=an/2ⁿ,数列{bn}是以1/2为首项,1为公差的等差数列(不是等比,是等差)
(2)
an/2ⁿ=1/2+1×(n-1)= n-1/2=(2n-1)/2
an=[(2n-1)/2]·2ⁿ=(2n-1)·2^(n-1)
n=1时,a1=(2×1-1)×1=1,同样满足通项公式
数列{an}的通项公式为an=(2n-1)·2^(n-1)

1.a_(1)=1,
a_(n+1)=2a_(n)+2^(n)----------------1
b_(n)=a_(n)/2^(n)
将式子1左右两边同时除以 2^(n+1),则:
b_(n+1)=b_(n)+1/2
b_(1)=a_(1)/2^(1)=1/2
所以数列{bn}是首项为1/2,公差为1/2的等差数列.
2.b_(n)=n/2.
a_(n)=b_(n)*2^(n)=n*2^(n-1)
数列{an}的通项为 a_(n)=n*2^(n-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
a(n+1)/2^n-an/2^(n-1)=1,为定值.
a1/2^(1-1)=1/1=1
数列{an/2^(n-1)}是以1为首项,1为公差的等差数列.
bn=an/2^(n-1)
数列{bn}是以1为首项,1为公差的等差数列.
an/2^(n-1)=1+(n-1)=n
an=n×2^(n-1)
Sn=a1+a2+...+an=1×2^0+2×2^1+...+n×2^(n-1)
2Sn=1×2^1+2×2^2+...+(n-1)×2^(n-1)+n×2^n
Sn-2Sn=-Sn=2^0+2^1+2^2+...+2^(n-1)-n×2^n
=(2^n-1)/(2-1)-n×2^n
=2^n-1-n×2^n
=(1-n)×2^n-1
Sn=(n-1)×2^n+1

1,A(n+1)=2An+2^n,
两边除以2^n得
A(n+1)/2^n=An/2^(n-1)+1,
即B(n+1)=Bn +1,
Bn是等差数列.
2,B1=A1=1,
则Bn=n,
即An=n2^(n-1)
Sn=1+2*2^1+3*2^2+.+n2^(n-1)
2Sn=2+2*2^2+3*2^3+.+n2^n,
相减得
Sn=n2^n-(1+2^1+2^2+...+2^(n-1))
=n2^n-(1-2^n)/(1-2)
=(n-1)2^n +1.

是一样的,按照公式：
1 + 2 + 2^2 + …… 2^(n-1)
=2^0+2^1+2^2+...+2^(n-1)
=(1-2^n)/(1-2)*(1)
=1*(2^n -1)/(2-1)
你写的式子只是把上下大的写在了前面,好看点,实际来说,就是：
(1-q^n)/(1-q)*a1
=(q^n-1)/(q-1)*a1