Question

等差数列A1=1,前 n项和满足S2n/Sn=4n+2/n+1 设Bn=（An）p^(An),求前n项和

Answer 1

因为S2n=2n(a1+a2n)/2=n[2a1+(2n-1)d] ,Sn=n(a1+an)/2=n[2a1+(n-1)d]/2
又S2n/Sn=4n+2/n+1，所以[2+(2n-1)d]/[2+(n-1)d]=(2n+1)/(n+1)对任意正整数n都成立，解得d=1,于是An=n,Bn=np^n，
(1)当p=1时，Bn前n项和为Tn=n(n+1)/2
(2)当p≠1时 ，B1=p^1, B2=2p^2 ,B3=3p^3 ,……，Bn=np^n ，相加得：
Tn=p^1+2p^2 +3p^3 +…+np^n, 乘以p得： pTn=p^2+2p^3 +3p^4 +…+np^(n+1)
错位相减得：（1-p)Tn=p^1+p^2 +p^3 +…+p^n-np^(n+1)
=p(1-p^n)/(1-p)-np^(n+1)
所以Tn=p(1-p^n)/(1-p)^2-np^(n+1)/(1-p)