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
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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
![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](/uploads/image/z/539850-66-0.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97A1%3D1%2C%E5%89%8D+n%E9%A1%B9%E5%92%8C%E6%BB%A1%E8%B6%B3S2n%2FSn%3D4n%2B2%2Fn%2B1+%E8%AE%BEBn%3D%EF%BC%88An%EF%BC%89p%5E%28An%29%2C%E6%B1%82%E5%89%8Dn%E9%A1%B9%E5%92%8C)
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)
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)
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