Question

数列｛an｝中，a1=1，an＋1=2an+2^n(1)设bn=an/2^n-1。证明数列{bn}是等差数列（2）求数列{an}的前n项和sn

Answer 1

bn = an/2^(n-1)
b<n-1> = a<n-1>/2^(n-2)
bn - b<n-1>
= an/2^(n-1) - a<n-1>/2^(n-2)
= (an - 2a<n-1> )/2^(n-1)
把 已知条件 a<n+1> = 2an+2^n 即 an = 2a<n-1> + 2^(n-1) 代入上式

bn - b<n-1>
= 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<n-1> + 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

Answer 2

1.a(n+1)=2an+2^n
a(n+1)/2^(n+1)=an/2^n+1/2
b(n+1)=bn+1/2
b1=a1/2-1=-1/2
{bn}是首项为-1/2，公差为1/2的等差数列
2.bn=-1/2+(n-1)/2=(n-2)/2=n/2-1
an=(bn+1)2^n
an=n/2*2^n
an=n*2^(n-1)
sn=a1+a2+...+an=1+4+..+n*2^(n-1)
2sn=2*a1+2*a2+...+2an=2+2*2^2+3*2^3+....+n*2^n
2sn-sn=n*2^n-n*2^(n-1)+(n-1)*2^(n-1)-(n-2)*2^(n-2)+...+1*2^1-1*2^0
sn=n*2^n-2^(n-1)-...-2^1-2^0
sn=n*2^n-2^n+1
sn=(n-1)*2^n+1

Answer 3

不说了提示很明显同时除以2^n得到：a(n+1）/2^n=an/2^(n-1)+1即b(n+1)=bn+1
bn=b1+(n-1）=n an=n*2^(n-1)
求和用错位相减Sn=1*1+2*2+3*4+……+n*2^(n-1)
2Sn=1*2+2*4+3*8+……（n-1)*2^(n-1)+n*2^n
两式相减-Sn=1+2+2^2+……2^(n-1)-n*2^n=(1-2^n)/(1-2)-n*2^n=2^n-1-n*2^n
Sn=(n-1)*2^n+1