已知数列{an}中,a1=3,a(n+1)=an+2^n-3n ,求数列{an}的通项公式。

请写出过程,谢谢
2025-03-07 01:22:34
推荐回答(1个)
回答1:

a(n+1)=an+2^n-3n
a(n+1)-an=2^n-3n
an-a(n-1)=2^(n-1)-3(n-1)
an - a1 = [2^1+2^2+...+2^(n-1) ] - [ 3+6+...+3(n-1)]
= 2^(n-1) -1 - 3n(n-1)/2
an = 2^(n-1) +2 - 3n(n-1)/2