Question

已知数列{An}的前n项和Sn满足Sn=1-2/3An.

求lim(A1S1+A2S2+…AnSn)。需要最详细的过程，谢谢!

Answer 1

n=1时 S1=1-(2/3)A1 解得A1=3/5
n>1时 S(n-1)=1-(2/3)A(n-1)
An=Sn-S(n-1)=-(2/3)An+(2/3)A(n-1)
所以An=(2/5)*A(n-1)
所以{An}是公比为2/5的等比数列
故An=(3/5)*(2/5)^(n-1)
所以Sn=1-(2/3)An
=1-(2/5)*(2/5)^(n-1)
=1-(2/5)^n
AnSn=An*[1-(2/3)An]
=An-(2/3)An²
=An-(2/3)*(3/5)²*(2/5)^(2n-2)
=An-(6/25)*(4/25)^(n-1)
所以A1S1+A2S2+...+AnSn
=(A1+A2+.....+An)-(6/25)*[1+(4/25)+...+(4/25)^(n-1)]
=Sn-(6/25)*[1-(4/25)^n]/(1-4/25)
=1-(2/5)^n-2/7+(2/7)*(4/25)^n
=5/7-(2/5)^n-(4/25)^n
故lim(A1S1+A2S2+…AnSn) (n→∞)
=5/7

Answer 2

Sn=1—2/3an,S(n-1)=1—2/3a(n-1)
Sn-S(n-1)=an=(1—2/3an)-(1—2/3a(n-1))
=2/3（a(n-1)-an)
an=2/5a(n-1)
an/a(n-1)=2/5
故{an}是首项a1=s1=1-2/3a1,a1=3/5，公比2/5的等比数列
an=3/5*(2/5)^(n-1)
sn=(3/5-3/5(2/5)^(n-1))/(1-(2/5))
=1-(2/5)^(n-1)
ansn=3/5(2/5)^(n-1)-3/5(2/5)^(2n-2)
a1s1+a2s2+.....ansn(分组求和法）
=1-（2/5）^(n-1）-5/7+5/7(4/25)^(n-1)
=2/7-(2/5)^(n-1)+5/7*(4/25)^(n-1)
取极限=2/7

Answer 3

略有点麻烦。把An通项公式算出来是一个等比数列，然后Sn=（1-2/3）An，乘进去变成An减去三分之二An平方。An平方仍然是等比数列，只是首项和公比变了。用极限分别解出来是1-(2/3)*(21/9)=1-2/7=5/7.答案七分之五