在数列{an}中,a1=1,且4a (n+1)-ana(n+1)+2an=9【括号内均为下标】,通过计算a2,a3,a4
2个回答
an= (6n-5)/(2n-1)
4a2-a2+2=9
a2=7/3
4a3-(7/3)a3+14/3=9
a3=13/5
4a4-(13/5)a4+26/5=9
a4=19/7
……
an= (6n-5)/(2n-1)
证明：对n进行归纳,设第n项成立
下面检验n+1项：
即看看4a（n+1）-anan+1+2an是否为9
：
因为：an=(6n-5)/(2n-1)
所以：a(n+1)=(6n+1)/(2n+1)
4a(n+1)-anan+1+2an
=4(6n+1)/(2n+1)-(6n-5)(6n+1)/(2n+1)(2n-1)+2(6n-5)/(2n-1)
=[4(6n+1)(2n-1)+2(6n-5)(2n+1)]/(2n+1)(2n-1)-(6n-5)(6n+1)/(2n+1)(2n-1)
=[4(6n+1)(2n-1)-(6n-5)(2n-1)]/(2n+1)(2n-1)
=[(2n-1)(24n+4-6n+5)]/(2n+1)(2n-1)
=(2n-1)(18n-9)/(2n+1)(2n-1)=9(2n-1)(2n+1)/(2n+1)(2n-1)
=9
证毕!
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