已知数列(an)满足a1=1/5,且当n>1,n属于正整数时,有an-1/an=(2an-1+1)/(1-2an)

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已知数列(an)满足a1=1/5,且当n>1,n属于正整数时,有an-1/an=(2an-1+1)/(1-2an)
(1)求证：数列(1/an)为等差数列
(2)试问a1a2是否是数列(an)中的项?如果是,是第几项；如果不是,请说明理由

a(1) = 1/5
a(n-1)/a(n) = [2a(n-1) + 1]/[1 - 2a(n)],
a(n) 不等于0或1/2.
a(n-1)[1 - 2a(n)] = a(n)[2a(n-1) + 1],
a(n-1) - 2a(n)a(n-1) = 2a(n)a(n-1) + a(n)
4a(n)a(n-1) = a(n-1) - a(n),
4 = 1/a(n) - 1/a(n-1),
所以,{1/a(n)}是首项为1/a(n) = 5,公差为4的等差数列.
1/a(n) = 5 + 4(n-1) = 4n + 1,
a(n) = 1/(4n+1),n = 1,2,...
a(2) = 1/(4*2+1) = 1/9,
a(1)a(2) = 1/5*(1/9) = 1/45 = 1/(4*11 + 1).
所以,
a(1)a(2)是数列{a(n)}中的项,是第11项.

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