1/2+2/2^2+3/2^3+.+n/2^n=2-(n+2)/2^n.
1、当n=1时候,
左边=1/2;
右边=2-3/2=1/2
左边=右边,成立.
2、设n=k时候,有:
1/2+2/2^2+3/2^3+.+k/2^k=2-(k+2)/2^k成立,
则当n=k+1时候:有:
1/2+2/2^2+3/2^3+.+k/2^k+(k+1)/2^(k+1)
=2-(k+2)/2^k+(k+1)/2^(k+1)
=2-[2(k+2)-(k+1)]/2^(k+1)
=2-(k+3)/2^(k+1)
=2-[(k+1)+2]/2^(k+1)
得证.