数列{an}的前n项和为Sn,a1=1,a2=2,an+2-an=1+(-1)n(n∈N+),则S100= ...
问题详情:数列{an}的前n项和为Sn,a1=1,a2=2,an+2-an=1+(-1)n(n∈N+),则S100=.【回答】由an+2-an=1+(-1)n知a2k+2-a2k=2,a2k+1-a2k-1=0,∴a1=a3=a5=…=a2n-1=1,数列{a2k}是等差数列,a2k=2k.∴S100=(a1+a3+a5+…+a99)+(a2+a4+a6+…+a100)=50+(2+4+6+…+100)=50+=2600.*:2600知识点...