Question

设数列{an}的前n项和为Sn=2n2,{bn}为等比数列,且a1=b1,b2(a2

设数列{an}的前n项和为Sn=2n2,{bn}为等比数列,且a1=b1,b2(a2-a1)=b1.求数列{an}和{bn}的通项公式.设cn=an除以bn求数列{cn}的前n项和Tn

Answer 1

n＞1
Sn-S(n-1)=an=4n-2
n=1
a1=S1=2
所以
an=4n-2
b1=a1=2
b2/b1=1/4
bn=2*(1/4)^(n-1)
cn=(2n-1)*4^(n-1)

Tn=1*4^0+3*4^1+5*4^2+......+(2n-1)*4^(n-1)
4Tn= 1*4^1+3*4^2+5*4^3+........... +(2n-1)*4^n
-3Tn=1+2*(4^1+4^2+.............4^(n-1))-(2n-1)*4^n
-3Tn=1+2*4(1-4^(n-1))/(1-4)-(2n-1)*4^n

Tn=5/9+(6n-5)*4^n/9