Question

数列0,1,1,2,2,3,3,4,4,...的前2n-1项和S2n-1=_____

Answer 1

数列0,1,1,2,2,3,3,4,4,...的前2n-1项和S2n-1=_____

a(2n-1)=n-1

a(2n)=n

所以S(2n-1)=a1+a2+a3+a4+...+a(2n-1)

=a1+a3+...+a(2n-1)+a2+a4+...+a(2n-2)

=(0+1+2+...+(n-1))+(1+2+...+(n-1))=n(n-1)