Question

高中数列题，求详细解答过程，谢谢！

Answer 1

解：
2an-a(n+1)=3ana(n+1)
等式两边同除以ana(n+1)
2/a(n+1) -1/an=3
2/a(n+1)=1/an +3
2/a(n+1)-6=1/an -3
2[1/a(n+1) -3]=1/an -3
[1/a(n+1) -3]/(1/an -3)=½，为定值
1/a1 -3=1/3 -3=-8/3
数列{1/an -3}是以-8/3为首项，½为公比的等比数列
1/an -3=(-8/3)·(½)ⁿ⁻¹=-8·(½)ⁿ⁻¹/3
1/an=3 -8·(½)ⁿ⁻¹/3=[9-8·(½)ⁿ⁻¹]/3
an=3/[9-8·(½)ⁿ⁻¹]=3·2ⁿ/(9·2ⁿ-16)
n=1时，a1=3·2/(9·2-16)=3，同样满足表达式
数列{an}的通项公式为an=3·2ⁿ/(9·2ⁿ-16)

Answer 2

1)
a1=3 , 3ana(n+1)=2an-a(n+1)
3ana(n+1)=2an-a(n+1)
等式两边除以ana(n+1)得
[2/a(n+1)]-[1/an]=3
[1/an]-3=2[1/a(n+1)-3]
[(1/an)-3]/[1/a(n+1)-3]=2
{(1/an)-3}是等比数列
(1/an)-3=[(1/a1)-3]*2^(n-1)=-8/3*2^(n-1)=-2^(n+2)/3
(1/an)-3=-2^(n+2)/3
1/an=[9-2^(n+2)]/3
an=3/[9-2^(n+2)]
2)
an=(2n+1)*(1/3)^n
Sn=3*1/3+5*(1/3)^2+7*(1/3)^3+...+(2n+1)*(1/3)^n
1/3Sn=3*(1/3)^2+5*(1/3)^3+...+(2n-1)(1/3)^n+(2n+1)*(1/3)^(n+1)
两式相减得
2/3Sn=1+2[(1/3)^2+(1/3)^3+...+(1/3)^n]-(2n+1)*(1/3)^(n+1)
2/3Sn=1+2{1/9[1-(1/3)^(n-1)]/(1-1/3)}-(2n+1)*(1/3)^(n+1)
2/3Sn=1+1/6[1-(1/3)^(n-1)]-(2n+1)*(1/3)^(n+1)
2/3Sn=7/6-[(4n+5)/6*(1/3)^n]
Sn=7/4-[3(4n+5)/4*(1/3)^n]
3)
an=3/[(6n-5)(6n+1)]=1/2*[1/(6n-5)-1/(6n+1)]
Sn=1/2[(1-1/7)+(1/7-1/13)+(1/13-1/19)+...+1/(6n-5)-1/(6n+1)]
Sn=1/2[1-1/(6n+1)]
Sn=3n/(6n+1)
4)
a1=3 , a(n+1)=3an-2
a(n+1)-1=3(an-1)
[a(n+1)-1]/(an-1)=3
an-1=(a1-1)*3^(n-1)=2*3^(n-1)
an=1+[2*3^(n-1)]

Answer 3

看不清楚啊