Question

数列{An}满足a1=1，且An=2An-1+2^n(n大于等于2且n属于整数).求数列的通项公式

如题

Answer 1

an/2^n
=(2an-1)/2^n + 1
=(an-1)/2^(n-1)+1
an/2^n - (an-1)/2^(n-1) = 1
则{an/2^n}是公差为1的等差数列。

设Tn=an/2^n
则Tn是公差为1的等差数列
Tn=T1 + (n-1)d=a1/2 + (n-1)*1=n-1/2
则an=(n-1/2)*2^n=n*2^n - 2^(n-1)

Answer 2

An/2^n=an-1/2^n-1+1,{An/2^n}是以1/2为首项，1为公差的等差数列，所以An/2^n=1/2+（n-1)
An=2^n(n-1/2)

Answer 3

an=2的（n-1)次方+2的2*（(n-1)次方