Question

已知数列{an}的前n项和为sn,且S2n-1=4n^2-2n+1 则 sn等于

Answer 1

换元法：
令m=2n-1则n=(m+1)/2
带入S(2n-1)=4n^2-2n+1可得:
S(m)=4*[(m+1)/2]^2-2*[(m+1)/2]+1
=m^2+m+1
所以S(n)=n^2+n+1