a(n+1)=an+1/2^n

a(n+1)=an+1/2^n
a2=a1+1/2
a3=a2+1/2^2
a4=a3+1/2^3
...
an=an-1+1/2^(n-1)
相加,有:
a2+a3+..an=a1+a2+..+an-1+(1/2+1/2^2+1/2^3+...+1/2^(n-1)
an=a1+(1/2(1-(1/2)^n)/(1-1/2)
=a1+(1/2)^n-1
a1没有给吗????
an=a1+(1/2)^n-1
a1=2
an=1+(1/2)^n

解：a2-a1=1/2
a3-a2=1/2^2
a4-a3=1/2^3
......
an-an-1=1/2^(n-1)
把上面n个式子相加抵消中间项得an-a1=1/2+...+1/2^(n-1)
=1/2(1-1/2^(n-1))/(1-1/2)=1-1/2^(n-1)
所以an=a1+1/2^(n-1)

a(n+1)=an+1/2^n
a(n+1)-an=1/2^n

由此知，bn=a(n+1)-an

对数列bn求和sn=（1-1/2^n）/(1/2)=2-2^(1-n)=an-a1

an=2-2^(1-n)+a1

因为a(n+1)=an+1/2^n

所以:
an=a(n-1)+1/2^(n-1)

a(n-1)=a(n-2)+1/2^(n-2)

..........

a2=a1+1/2

以上式子叠加:
an=1/2^(n-1)+1/2^(n-2)+......+1/2

=a1+(1/2(1-(1/2)^n)/(1-1/2)

=a1+(1/2)^n-1

只须把a1代入即可

要先知道a1才能求的.
a(n+1)=