Question

求an,an满足a1等于1,an+1-an=2的n次方+1

设数列｛an｝满足a1=1,an+1-an=2的n次方加1（n∈正整数）（1）求数列｛an｝的通项公式；（2）求数列｛an｝的前n项和Sn

Answer 1

简单分析一下，详情如图所示

Answer 2

(2)
a(n+1)-an =2^n +1
an-a(n-1) =2^(n-1) +1
an - a1 = [2+2^2+...+2^(n-1)] + (n-1)
an = 2(2^(n-1) -1) + n
= n-2 + 2^n
(2)
Sn = a1+a2+...+an
= n(n+1) - 2n + 2(2^n -1)
=n^2-n-2 + 2^(n+1)