第n项:n又1/(2^n)
1又1/2+2又1/4+3又1/8+4又1/16+n又1/(2^n)
=(1+2+3+...+n)+(1/2+1/4+1/8+1/16+...+1/2^n)
=[(n+1)n/2]+(1-1/2^n)
=n^2/2+n/2+1-1/2^n