已知数列An的前N项和为Sn=2n^2bn为等比数列且b1=a1b4=1/32设Cn=An/bn求Cn的前N项和Tn

Question

已知数列An的前N项和为Sn=2n^2 bn为等比数列 且b1=a1 b4=1/32 设Cn=An/bn 求Cn的前N项和Tn

Answer 1

n=1时,a1=S1=2×1²=2
n≥2时,Sn=2n² S(n-1)=2(n-1)²
an=Sn-S(n-1)=2n²-2(n-1)²=4n-2
n=1时,a1=4-2=2,同样满足通项公式.
数列{an}的通项公式为an=4n-2
设{bn}公比为q
b4/b1=q³=(1/32)/2=1/64
q=1/4
bn=b1q^(n-1)=2×(1/4)^(n-1)=8/4ⁿ
cn=an/bn=(4n-2)/(8/4ⁿ)=(2n-1)×4^(n-1)
Tn=c1+c2+...+cn=1×1+3×4+5×4²+...+(2n-1)×4^(n-1)
4Tn=1×4+3×4²+...+(2n-3)×4^(n-1)+(2n-1)×4ⁿ
Tn-4Tn=-3Tn=1+2×4+2×4²+...+2×4^(n-1) -(2n-1)×4ⁿ
=2×[1+4+4²+...+4^(n-1)] -1 -(2n-1)×4ⁿ
=2×1×(4ⁿ-1)/(4-1) -1 -(2n-1)×4ⁿ
=(5-6n)×4ⁿ/3 -5/3
Tn=(6n-5)×4ⁿ/9 +5/9