Question

数列1又1/2,2又1/4,3又1/8,4又1/16,……的前n项和

Answer 1

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