Question

高一数学问题 关于数列

在数列an中,a1=1 an+1=(1+1/n)an+(n+1)/2^n
求an的通项公式

Answer 1

an+1=(1+1/n)an+(n+1)/2^n
两边同除以（n+1）得到：a（n+1）/（n+1）=an/n+1/2^n
设bn=an/n,b1=a1/1=1
所以b（n+1）=bn+1/2^n
b2=b1+1/2
b3=b2+1/2^2
...
bn=b(n-1)+1/2^(n-1)
各项相加得:
bn=b1+(1/2+1/2^2+....+1/2^(n-1))=1+1/2*(1-1/2^(n-1))/(1-1/2)=1+1-2/2^n
所以bn的通项公式是：bn=2-1/2^n
an的通项公式是:an=nbn=n(2-1/2^n)

Answer 2

an+1=an+(n+1)
则有:
an=a(n-1)+n
a(n-1)=a(n-2)+(n-1)
-------
a2=a1+2
a1=a1
左右相加得:
a1+a2+a3+-----an=a1+a1+a2+a3+-----+a(n-1)+2+3+4+-----+n
an=a1+2+3+4+---+n
=(n^2+n+2)/2

Answer 3

a(n+1) / (n+1) = an / n + 1/2^n
设b(n)=an / n
b(n+1)=bn + 1/2^n
b1=a1/1=1
b(n)=1+1/2+1/4+1/8+...+1/2^(n-1)
=2-1/2^(n-1)
所以an = n*bn = 2n - n/2^(n-1)

Answer 4

an+1=[(n+1)/n]an+(n+1)/2^n
两边同时除以（n+1),得
[an+1/(n+1)]=(an/n)+(1/2^n)
所以就有an/n-[an-1/(n-1)]=[1/2^(n-1)]
[an-1/(n-1)]-[an-2/(n-2)]=[1/2^(n-2)]
.......
a2/2-a1/1=1/2^1
左边与右边对应相加
an/n-a1/1=1/2^(n-1)+1/2^(n-2)+……+1/2^1
an/n-1=1-1/2^(n-1)
an/n=2-1/2^(n-1)
an=2n-n/2^(n-1)