Question

数列an满足a1=2,an-an-1=1/2^n

已知数列{an}满足；a1=1/2,且an-an－1=1/2^n(1)求数列{an}的通项an(2)求前n项和Sn 注意：n-1是n的下标
求具体过程 谢谢!!

Answer 1

（1）an-an－1=1/2^n
a(n-1)-a(n-2)=1/2^(n-1)
┄┄
a2-a1=1/2²
上式相加得：an-a1=1/2^n+1/2^(n-1)┄┄+1/2²
an=1/2^n+1/2^(n-1)┄┄+1/2²+1/2=1-1/2^n

（2）Sn=a1+a2+....an=n-1+1/2^n (n>=2)
S1=1/2 符合上式
所以Sn=a1+a2+....an=n-1+1/2^n

Answer 2

an-an－1=1/2^n
a(n-1)-a(n-2)=1/2^(n-1)
┄┄
a2-a1=1/2²
上式相加得：an-a1=1/2^n+1/2^(n-1)┄┄+1/2²
an=1/2^n+1/2^(n-1)┄┄+1/2²+1/2=1-1/2^n
Sn=n-1+1/2^n。