已知数列an的前n项和为Sn=n²+2n-1,则a1+a3+a5+…………a24+a25=?

已知数列an的前n项和为Sn=n²+2n-1,则a1+a3+a5+…………a24+a25=?

n=1,a1=S1=2
n≥2,an=Sn-S(n-1)=(n²+2n-1)-[(n-1)²+2(n-1)-1]
=(n²+2n-1)-(n²-2)
=2n+1
所以 a1+a3+a5+……a25
= 2+(7+11+.+51)
=2 +(7+51)*12/2
=2+58*6
=350