Question

求数列前n项和：1+2*3+3*7+4*15+...+n(2^n-1)

Answer 1

第n项=1+2+……n=n(n+1)/2=n^2/2+n/2

所以Sn=(1^2+2^2+……+n^2)/2+(1+2+……n)/2

=[n(n+1)(2n+1)/6]/2+[n(n+1)/2]/2

=n(n+1)(n+2)/6