Question

数列｛an｝的前项和记为sn对任意的正整数n，均有4Sn=(an+1)²,且an>0 (1)求a1

数列｛an｝的前项和记为sn对任意的正整数n，均有4Sn=(an+1)²,且an>0
(1)求a1,a2的值(2)求数列｛an｝的通项公式（3）若bn=an/3n(n∈N*),求数列｛bn｝的前n项和Tn

Answer 1

4Sn=(an+1)²
4a1=4S1=(a1+1)^2
a1^2-2a1+1=0
a1=1
4S2=4(a1+a2)=(a2+1)^2
4(a2+1)=(a2+1)^2
a2^2-2a1-3=0
a2=3或a2=-1(舍去)
所以,a1=1 ,a2=3
2)4Sn-4S(n-1)=(an+1)²-[a(n-1)+1]²
4an=(an+1)²-[a(n-1)+1]²
(an-1)²-[a(n-1)+1]²=0
[(an-1+a(n-1)+1][(an-1)-[a(n-1)+1]=0
[an+a(n-1)][an-[a(n-1)-2]=0
an-a(n-1)=2
an=a1+2(n-1)=2n-1
an=2n-1
3)bn=an/3n,题有错????

追问

哪里

3.若bn=a^n/3^n(n∈N*)……

追答

bn=a^n/3^n=(2n-1)/3^n
Tn=1/3+3*(1/3)^2+5*(1/3)^3+..+(2n-3)*(1/3)^(n-1)+(2n-1)*(1/3)^n
1/3Tn= (1/3)^2+3*(1/3)^3+..+(2n-5)*(1/3)^n+(2n-1)*(1/3)^(n+1)
两式相减得
2/3Tn=1/3+2[(1/3)^2+3(1/3)^3+..+(1/3)^n]-(2n-1)*(1/3)^(n+1)
=1/3+2[1/6(1-(1/3)^(n-2)]-(2n-1)*(1/3)^(n+1)=2/3-(2n+8)(1/3)^(n+1)
Tn=1-[(2n+4)*(1/3)^n]

Answer 2

S1=a1

a1=1

Answer 3