數列{an}的前n項和為Sn,a1=1,a2=2,an+2-an=1+(-1)n(n∈N+),則S100= ...
問題詳情:數列{an}的前n項和為Sn,a1=1,a2=2,an+2-an=1+(-1)n(n∈N+),則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,數列{a2k}是等差數列,a2k=2k.∴S100=(a1+a3+a5+…+a99)+(a2+a4+a6+…+a100)=50+(2+4+6+…+100)=50+=2600.*:2600知識點...