Sn=1+(1+1⼀2)+(1+1⼀2+1⼀4)+....+[1+1⼀2+1⼀4.....+1⼀2^(n-1)]=?

过程详细点
2025-02-23 22:08:53
推荐回答(1个)
回答1:

an = 1+1/2+1/4.....+1/2^(n-1) = 2 - 1/2^(n-1)
所以Sn = 2 - 1 + 2 - 1/2 + 2 - 1/4 + ... + 2 - 1/2^(n-1)
= 2n - (1+1/2+1/4+...+1/2^(n-1) )
= 2n - (2 - 1/2^(n-1))
= 2n - 2 + 1/2^(n-1)