数列推导设等差数列An的前n项和为Sn,则S4,S8-S4,S12-S8,S16-S12成等差...答:a6=a2=4d...a(n+4)=a(n)+4d (S8-S4)-S4=4d [S(4n+4)-S(4n)]=4d 同理可得S4,S8-S4,S12-S8,S16-S12成等差数列 t5=t1*q^4,t6=t2*q^4...t(n+4)=t(n)*q^4 (T8/T4)/T4=q^4 [T(4n+4)/T(4n)]=q^4 同理可得T4,T8/T4,T12/T8,T16/T12成等比数列 ...