大师作文网数列An满足 根号下AnAn+1 +AnAn+2=4根号下AnAn+1 +An+1An+1 +3根号下AnAn+1 且A
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数列An满足 根号下AnAn+1 +AnAn+2=4根号下AnAn+1 +An+1An+1 +3根号下AnAn+1 且A1=1,A2=8求An
这道题,你要有点儿耐心看下去!
√[ ],表示[ ]内都在平方根号下
√[ana(n+1)+ana(n+2)]
=4√[ana(n+1)+a(n+1)a(n+1)]+3√[ana(n+1)]
√an*√[a(n+1)+a(n+2)]=4√a(n+1)*√[an+a(n+1)]+3√[ana(n+1)]
两边同除以√an*√a(n+1),√[a(n+1)+a(n+2)]/√a(n+1)=4√{[an+a(n+1)]/√an}+3
√[1+ a(n+2)/ a(n+1)]=4√[1+ a(n+1)/an]+4-1,设bn=a(n+1)/an
1+√(1+b(n+1)=4[1+√(1+bn)],设cn=1+√(1+bn)
c(n+1)=4cn,c1=1+√(1+b1)=1+√(1+a2/a1)=1+√(1+8/1)=4,
cn=4*4^(n-1)=4^n
1+√(1+bn)=4^n,bn=(4^n-1)^2-1=4^n(4^n-2)
an=a2/a1*a3/a2*……*an/a(n-1)
=4*4^2*……4^(n-1)*(4-2)*(4^2-2)*……*(4^(n-1)-2)
=4^(n(n-1)/2)* (4-2)*(4^2-2)*……(4^(n-1)-2)
√[ ],表示[ ]内都在平方根号下
√[ana(n+1)+ana(n+2)]
=4√[ana(n+1)+a(n+1)a(n+1)]+3√[ana(n+1)]
√an*√[a(n+1)+a(n+2)]=4√a(n+1)*√[an+a(n+1)]+3√[ana(n+1)]
两边同除以√an*√a(n+1),√[a(n+1)+a(n+2)]/√a(n+1)=4√{[an+a(n+1)]/√an}+3
√[1+ a(n+2)/ a(n+1)]=4√[1+ a(n+1)/an]+4-1,设bn=a(n+1)/an
1+√(1+b(n+1)=4[1+√(1+bn)],设cn=1+√(1+bn)
c(n+1)=4cn,c1=1+√(1+b1)=1+√(1+a2/a1)=1+√(1+8/1)=4,
cn=4*4^(n-1)=4^n
1+√(1+bn)=4^n,bn=(4^n-1)^2-1=4^n(4^n-2)
an=a2/a1*a3/a2*……*an/a(n-1)
=4*4^2*……4^(n-1)*(4-2)*(4^2-2)*……*(4^(n-1)-2)
=4^(n(n-1)/2)* (4-2)*(4^2-2)*……(4^(n-1)-2)
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