Question

已知数列An满足，A1=1，An+1=An+n+2*n,求An

Answer 1

不知道是不是An+1=An+n+2^n
令An=Bn+Cn; Bn+1=Bn+n; Cn+1=Cn+2^n; B1=0, C1=1
这样An+1=Bn+1+Cn+1=Bn+Cn+n+2^n=An+n+2^n，同时A1=1依然成立
可以求出Bn=n(n-1)/2; Cn=-1+2^n
得到An=n(n-1)/2-1+2^n

Answer 2

a(n+1)=a(n)+n+2^n
a(n+1)-2^(n+1)=a(n)-2^n +n
a(n+1)-2^(n+1)=a(n)-2^n + (1/2)[(n+1)^2-n^2] - (2n+1)/2 + n
a(n+1)-2^(n+1) - (1/2)(n+1)^2 = a(n) - 2^n - (1/2)n^2 - 1/2
{a(n)-2^n - (1/2)n^2}是首项为a(1)-2-(1/2)=-3/2,公差为(-1/2)的等差数列.
a(n)-2^n-(1/2)n^2 = -3/2 -(n-1)/2 = - 1 - n/2
a(n)=2^n+(1/2)n^2 - 1 - n/2 = 2^n + (n^2 - 2n - 2)/2