1⼀2+2⼀2^2+3⼀2^3+...+n⼀2^n=?

求详细解法
2024-12-02 21:30:39
推荐回答(1个)
回答1:

1/2+2/2^2+3/2^3+...+n/2^n
=(1/2+2/2^2+3/2^3+...+n/2^n)*2-(1/2+2/2^2+3/2^3+...+n/2^n)
=1+2/2+3/2^2+4/2^3+...+n/2^(n-1)-(1/2+2/2^2+3/2^3+...+n/2^n)
=1+1/2+1/4+......1/2^(n-1)-n/2^n
=2-1/2^(n-1)-n/2^n
=1又(2^n-2-n)/2^n